vedanta Excel In Opt. Mathematics - Book 10
Now,
coefficient of x coefficient of y constant
344
2 5 -7
Now, D = 3 4 = 15 – 8 = 7
2 5
4 4
D1 = –7 5 = 20 + 28 = 48
D2 = 3 4 = –21 – 8 = –29
2 –7
By using Cramer's rule,
x= DD1 = 48
7
D2
y= D = –29
7
48 -29
? x= 7 and y = 7
Example 5: A marketing boy sales 10 kg of rice, 12 kg of tea at cost Rs 1460 on Sunday.
Solution: On Monday he sales 5 kg of rice and 7 kg of sugar at cost Rs 810. Find the
costs of each kg of rice and sugar. (Use Cramer's rule)
Let Rs x and Rs y be the prices of per kg of rice and sugar respectively.
Then, we have,
10x + 12y = 1460
or, 5x + 6y = 730 ........ (i)
and 5x + 7y = 810 ........ (ii)
Now, to solve above equations using Cramer's rule,
coefficient of x coefficient of y constant
5 6 730
5 7 810
Now, D = 5 6 = 35 – 30 = 5
5 7
730 6
D1 = 810 7 = 5110 – 4860 = 250
D2 = 5 730 = 4050 – 3650 = 400
5 810
By using Cramer's rule,
x= DD1 = 250 = 50
5
D2 400
y= D = 5 = 80
? The costs of per kg of rice and sugar are Rs 50 and Rs 80 respectively.
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Exercise 7.4
Very Short Questions :
1. (a) In equations a1x + b1y = c1 and a2x + b2y = C2, what are the determinants
representing D, Dx and Dy?
(b) If D = 20, Dx = 20, Dy = 40, write the values of x and y.
(c) If D = 4, Dx = 1 , Dy = 1 , find the values of x and y.
2 4
Long Questions :
2. Solve the following system of equations by using Cramer's rule:
(a) 3x – 2y = 1 and –x + 4y = 3
(b) 3x – 5y = 2 and 2x + y = 4
(c) 2x – 3y = 2 and 4x – y = 1
(d) 5x – 4y = –3 and 7x + 2y = 6
(e) 2x + 3y = 17 and 9x – 8y = 12
(f) 2x – y = 1 and 3x + y = 9
(g) 2x – 3y = 3 and 4x – y = 11
(h) 3x + 5y = 21 and 2x + 3y = 13
(i) 5x + 8y = 340 and 7x + 6y = 320
(j) 5x – 3y = 8 and 9x + 7y = 126
3. Solve the following system of equations by using Cramer's rule:
(a) 3 + 2 = 1 and 4 + 3 = 17
x y x y 6
3 5 4 3 29
(b) x + y = 1 and x + y = 30
(c) x – y = 2 and x + y = 59
3 4 8 3 24
2 5 3 6
(d) x + y = 30 and x + y = 27
(e) x – y = –6 and 3x – 1 = y
6 4
4. Solve the following equations by Cramer's rule:
(a) 2x – 5 = 28 and 4x + 3 = –9
y y
10 4
(b) x – 2y = –1 and x + 3y = 11
(c) 8 – 9 = 1 and 10 + 6 = 7
x y x y
(d) 2(3x – y) = 5(x – 2) and 3(x + 4y) = 2(y – 3)
(e) 7(x – y) = x + y and 5(x + y) = 35 (x – y)
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5. A helicopter has 4 seats for passengers. Those willing to pay first class fares can take 60
kg of baggage each, but tourist class passengers are restricted to 20 kg each. It can carry
only 120 kg of baggage altogether. To find the number of passengers of each kind, use
Cramer's rule.
6. If the total cost of 6 kg potatoes and 4 kg of tomatoes is Rs. 200 and the total cost of 1 kg
of potatoes and 1 kg tomatoes is 42. Find the price of per kg of each vegetable by using
determinant.
Project Work
7. Write two simultaneous equations related to cost of two daily used commodities and
solve them by using Cramer's rule.
1. (a) D = aa12 bb12 , Dx = cc12 bb21 , Dy = aa12 cc12 (b) (1, 2) (c) 1 , 1
8 16
2. (a) (1, 1) (b) 22 , 8 (c) 1 , - 3 (d) 9 , 51
13 13 10 5 19 38
(e) (4, 3) (f) (2, 3) (g) (3, 1) (h) (2, 3)
(i) (20, 30) (j) (7, 9) 3.(a) - 3 , 2 (b) (6, 10)
8 9
1 1
(c) (9, 4) (d) - 15 , 12 (e) (12, 8)
4. (a) 3 , - 1 (b) (2, 3) (c) (2, 3) (d) -7, 3
2 5 6. (16, 26) 2
(e) No solution 5. (1, 3)
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Co-ordinate Geometry
8.0 Review B(x2,y2)
Let us review some formulae of coordinate geometry which we studied
in previous class.
(i) Distance Formula :
The distance between two points A (x1, y1) and B(x2, y2) is A(x1,y1)
given by
d = AB = (x2 - x1)2 + (y2 - y1)2 B(x2, y2)
(ii) Section Formula:
m2
(a) Internal Division : If P divides the join of A(x1, y1) and m1 P(x, y)
B(x2, y2), in ratio of m1 : m2 internally, then, coordinates
of P are given by
( (P(x,y) = P A(x1, y1)
m1x2 +m2x1 , m1y2 +m2y1
m1 + m2 m1 + m2
( (If the ratio is k:1, then, P(x,y) = P
kx2 +x1 , ky2 +y1
k+ 1 k+ 1
(b) External Division : If P divides the joining of A(x1, y1) and B(x2, y2) in m1 : m2 ratio,
externally, then, the coordinates of P are given by,
( (P (x, y) = P P(x , y)
m1x2 - m2x1 , m1y2 - m2y1 m2
m1 - m2 m1 - m2
If the ratio is k:1, them m1 B(x1, y1)
( (P (x, y) = P
kx2 - x1 , ky2 - y1
k-1 k-1
(c) Mid point formula : If P divides AB in two equal A(x1, y1)
parts, then, P is the midpoint of AB, the coordinates of P are given by,
( (P(x, y) = P
x1 + x2 , y1 + y2 A(x1, y1) P(x, y) B(x2, y2)
2 2
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(iii) Slope formula : The slope of line segment joining two B(x2,y2)
points A(x1, y1) and B(x2,y2) is given by m = y2 - y1
If AB makes an angle T with positive X-axis, x2 - x1
A(x1,y1) A(x1, y1)
then (m) = tanT F 2E
G
(iv) Centroid of triangle : The point of intersection of B(x2, y2) 1
medians of triangle is called centroid. In the given D C(x3, y3)
figure,G is the centroid of triangle
Y P(x, y)
x1 + x2 + x3 y1 + y2 + y3
G(x,y) = G 3 , 3
(vi) Equation of straight line in slope- intercept. BT
c
In the figure, straight line AB has, y-intercept = OB = c X T X’
AO
Y’
Equation of the line is y = mx + c. Q Y
B(0, b)
(vii) Equation of straight line in double intercept form.
b
x-intercept of PQ =OA = a A(a, 0)
X
y - intercept of PQ = OB = b X' O a P
Equation of PQ is x + y = 1. Y' Y
a b M B
(viii) Equation of straight line in normal form P
p
perpendicular distance of MN from origin = OP = p
X' O D A X
Angle made by OP with OX is AOP = D Y' N
Equation of straight line MN is x cosD + y sinD = p
Y N
(ix) Equations of straight line in point slope form P (x1, y1)
slope of MN = (m) = tanT
MN passes through the point P(x1, y1) X' T X
Equation of MN is y - y1 = m (x - x1) N
O
M
Y'
Y
(x) Equation of straight line in two points form Q(x2, y2)
P (x1, y1)
Equation of straight line joining two points P(x, y) and X'
Q(x2, y2) is M OX
Y'
y - y1 = y2 - y1 (x - x1)
x2 - x1 155
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(xi) Perpendicular distance of a point from a line. P(x1, y1)
The perpendicular distance of the P(x1, y1) from the line A Q B
AB with equation ax + by+ c = 0 is d = ax1 + by1 + c straight line of the form
a2 + b2
(xii) Discuss how to change the linear equation of
ax + by + c = 0 into following three standard forms:
(a) y = mx + c (b) x + y = 1 (c) xcosD + ysinD = p
a b
(xiii) Area of triangle : Let A(x1, y1), B(x2, y2) and C(x3, y3), be three vertices of ∆ABC. Then area
of ∆ABC is given by,
∆= 1 |x1y2 - x2y1+ (x2y3 - x3y2) + (x3y1- x1y3)|
2
8.1 Angle between two straight lines
Let us consider the following three case of two straight line l1 and l2.
(i) l1
O
T
l2
Here, two straight lines l1 and l2 intersect at O and four angles are formed.
(ii) l1
O l2
Here, two straight lines l1 and l2 are perpendicular to each other.
(iii) l1
l2
Two lines l1 and l2 are parallel to each other. Y
The angle between them is zero. QN
To find the angle between two lines y = m1x + c1 y = m2x + c2 S S–T
T y = m1x + c1
and y = m2x + c2
Let MN and PQ be two straight lines with equations X' T T X
y = m1x + c1 and y = m2x + c2. OA B
Let the lines intersect at S. The lines MN and PQ cut MP
OX at A and B respectively making angles T2, and T
with the positive direction of X-axis. Y'
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Then, slope of line MN (m1) = tanT, slope of line PQ (m2) = tanT2
Let the one angle between MN and PQ be T, i.e. and MSP = T
Then, by plane geometry,
T1 = T + T2 [ ? Exterior angle of a triangle is sum of two opposite interior angles]
or, T = T1 - T2
? tanT = tan(T1 - T2)
tanT1 - tanT2
? tanT = 1 + tanT1.tanT2m1 - m2
( (MSP = T = tan-1 1 + m1. m2 ...... (i)
Also, PSN is angle between the lines
PSN = 180° - T
? tan (PSN) = tan (180° - T) = - tanT = - m1 - m2
m1 - m2
( ( PSN = tan-1 - 1 + m1.m2 ....... (ii) 1 + m1.m2
combining above (i) and (ii), the angle between
( (m1 - m2
MN and PQ = tan-1 ± 1 + m1.m2
This relation gives an angle between the two lines. Another angle is its supplement.
m1 - m2
If 1 + m1.m2 is positive, the acute angle between the two lines is obtained. Otherwise
it gives the obtuse angle.
?
( (Note :
T = tan-1 m1 - m2
± 1 + m1.m2
(a) If the value of tanT is positive, T is acute angle.
(b) If the value of tanT is negative, T is obtuse angle.
(c) If the value of tanT is zero, the lines is parallel or coincident.
(d) If the value of tanT is undefined (f) when the value of T is 90°.
To find the condition that two lines y = m1x + c1 and y = m2x + c2 are parallel.
If two lines y = m1x + c1 and y = m2x + c2 are parallel, to each other then angle between
them is °.
i.e. T = 0°
or, 0ta=n01° m=+11m-m+m11m1m-2m1m1 2m1 - m2 = 0
? m1 = m2
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Therefore, two lines are parallel if their slopes are equal or if the lines are parallel their
slopes are equal.
To find the condition that two lines y = m1x + c1 and y = m2x + c2 are perpendicular
If two lines are perpendicular, the angle between them is 90°
i.e, T = 90°
or, tcsaoinns999000°°°==1m1+m1+m-1 mm-1mm11m21 2 or, 1 = m1 - m1
or, 1 + m1 . m2 = 0 0 1+ m1m2
or, m1 . m2 = –1
or, m1m2 = –1,
Therefore, two lines are at right angles if the product of their slopes is equal to –1.
To find the angle the lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Equations of two lines are:
a1x + b1y + c1 = 0......... (i)
a2x + b2y + c2 = 0 ........ (ii)
slope of line (i), m1 = – a1
slope of line (ii), m2 = - ba12
b2
Let T be the angle between the two lines,
( ( ( ((( ((tanTm1 - m2 - a1 – - a2
+ m1.m2 b1 b2
then = ± 1 =± a1 a2
? b1 b2
1+ - -
( ( ( (= ±
a1b2 – a2b1 =± a1b2 – a2b1
b1b2 + a1a2 a1a2 + b1b2
( (T tan-1
± a1b2 - a2b1
a1a2 + b1b2
( (Therefore, the angle between two lines is tan–1
± a1b2 - a2b1 .
a1a2 + b1b2
To find the condition for parallelism of two lines a1x + b1y + c1 = 0
and a2x + b2y +c2 = 0
a1b2 - a2b1
We have, tanT = a1a2 + b1b2
For parallelism, T = 0°, tan0° = a1b2 - a2b1
a1a2 + b1b2
a1 b1
or, a1b2 - a2b1 = 0 ? a2 = b2
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or, a1b2 = a2b1
Therefore, the two lines are parallel if the coefficients of x and y are proportional.
To find the condition for perpendicularity or orthogonality of two lines
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0.
a1b2 - a2b1
We have, tanT = a1a2 + b1b2
cotT = a1a2 + b1b2
a2b1 - a1b2
If T = 90° then cot90° = 0
or, cot90° = a1a2 + b1b2 or, 0= a1a2 + b1b2
a2b1 - a1b2 a2b1 - a1b2
? a1a2 + b1b2 = 0
Therefore, two lines are at right angles if the sum of the product of coefficients of x and
product of the coefficient of y is 0.
To find the equation of a line parallel to ax + by + c = 0
The given equation of line is ax + by + c = 0 ......... (i)
Slope of line (i), m1 = - coefficient of x = - a
coefficient of y b
Let the equation of a line parallel to (i) be y = mx + c ........ (ii)
Slope of line (ii), m2 = m
Since two lines (i) and (ii) are parallel m1 = m2
or, a = m
b
Putting the value of m in equation (ii),
y = - a x + C
b
or, by + ax + (-bc) = 0
or, ax + by + k = 0, where k = -bc.
Hence, the required equation of straight line parallel to (i) is ax + by + k = 0
Therefore, we change only the constant term in the given equation to get the equation of any
line parallel to given line.
Equation of a straight line perpendicular or orthogonal to ax + by + c = 0.
The equation of given line is ax + by + c = 0......... (i)
Its slope (m1) = - coefficient of x =- a
coefficient of y b
Let the equation of a straight line perpendicular to (i) be y = mx + c........ (ii)
Its slope (m2) = m
Since two lines (i) and (ii) are perpendicular m1.m2 = -1
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( (or, - a m=- 1 ? m = b
b of m in equation y= a
Putting the b
value (ii), we get a x+c
or, ay = bx + ac
or, bx - ay + ac = 0
or, bx - ay + k = 0 Where k = ac
Therefore, the required equation of straight line perpendicular to (i) is bx - ay + k = 0.
Therefore, we interchange the coefficients of x and y in given equation and sign of any one
of them is also changed. The constant term of given equation is also changed.
Worked out Examples
Example 1. Find the acute angle between the lines:
Solution:
y - (2 + 3 ) x - 5 = 0 ................(i)
Example 2.
Solution: y - (2 - 3 ) x - 2 = 0 ................. (ii)
160 Slope of line (i) m1 = - coefficient of x
coefficient of y
= (-) (2 + 3 ) =2+ 3
1 3
coefficient of x
Slope of line (ii), m2 = - coefficient of y = 2 -
Let T be the angle between given lines. Then
tanT = ± (m1 - m2)
1+ m1m2
2+ 3 -2+ 3
= 1+ (2 + 3 ) (2 - 3 )
= ± 2 3 =± 2 3 =± 3
1+ 4 -3 2
For acute angle we take positive sign only, we get
tanT = 3
or, tanT = tan60°
? T = 60°
Therefore, the required angle is 60°.
Find the obtuse angle between the given lines. ( i.e., y = mx form)
y = 3 x and 3 x + y + 3 = 0
Given equations of the lines are,
y = 3 x ........(i)
3 x + y + 3 = 0.......... (ii)
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slope of line (i), m1 = 3
slope of line (ii), m2 = - 3 =- 3
1
Let T be the angle between the given lines. Then
tanT = ± (m1 - m2) =± 3 +3 3) =± 2 3
1+ m1m2 1+( 3 ) (- 1-3
=±2 3 =±( 3)
-2
For obtuse angle, we take negative sign,
tanT = - 3
or, tanT = tan 120° ? T =120°
Therefore, the required angle is 120°
Example 3. Show that the lines 4x - 8y + 7 = 0 and x - 2y + 8 = 0 are parallel to each
Solution: other.
Example 4. The given equations of lines are 4x - 8y + 7 = 0 ......... (i)
Solution:
x - 2y + 8 = 0 ......... (ii)
Example 5.
Solution: Slope of line (i), m1 = - coefficient of x = - 4 = 1
coefficient of y -8 2
Slope of line (ii), m2 = - coefficient of x = - 1 = 1
coefficient of y -2 2
? m1 = m2
Therefore, the given lines are parallel to each other.
Show that the lines 2x + y + 2 = 0 and x - 2y + 3 = 0 are perpendicular to
each other.
The equations of given lines are
2x + y + 2 = 0 ........ (i)
x - 2y + 3 = 0 ........... (ii)
Slope of line (i), m1 = - coefficient of x = – 2 = –2
coefficient of y 1
Slope of line (ii), m2 = - coefficient of x = – 1 = 1
coefficient of y -2 2
1
Here, m1.m2 = - 2 × 2 =-1
Therefore, the lines (i) and (ii) are perpendicular to each other.
If the lines 2x + 3y = 5 and kx + y = 2 are parallel to each other. Find the
value of k.
The equation of given lines are :
2x + 3y = 5 ............ (i)
kx + y = 2 ............ (ii)
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slope of line (i), m1 = - coefficient of x = 2
coefficient of y 3
coefficient of x k
slope of line (ii), m2 = - coefficient of y = – 1 =-k
Since, the lines (i) and (ii) are parallel, we have
m1 = m2
or, - 2 =-k
3
2
or, k = 3
Therefore, the value of k is 2 .
3
Example 6. If the lines 3x + 4y - 7 = 0 and kx - 3y + 5 = 0 are perpendicular to each
Solution: other, find the value of k.
Example 7. The equations of given lines are,
Solution:
3x + 4y + 7 = 0 .......(i)
Example 8.
Solution: kx - 3y + 5 = 0 ......... (ii)
slope of line (i), m1 = - coefficient of x =- 3
coefficient of y 4
coefficient of x k k
slope of line (ii), m2 = - coefficient of y = - -3 = 3
since the lines (i) and (ii) are perpendicular to each other, m1.m2 = -1
or, - 3 × k =-1 or, k = 4
4 3
Therefore, the value of k is 4.
If the line joining the points (-1, k), and (5,6) is perpendicular to
2x + y + 7 = 0, find the value of k.
The equation of given line is 2x + y + 7 = 0
Its slope (m1) = - coefficient of x = - 2 = -2
coefficient of y 1
slope of line joining points (-1, k) and (5,6)
(m2) = y2 - y1 = 6-k = 6-k
x2 - x1 5+1 6
Since the lines are perpendicular, m1.m2 = -1
or, -2 × 6-k =-1 or, 6 - k = 0
2
? k=6
Find the equation of straight line which is parallel to the line
4x -3y + 12 = 0 and passing through the point (1, 2).
Equation of given straight line is 4x - 3y + 12 = 0......... (i)
Its slope is (m1) = - coefficient of x = - 4 = 4
coefficient of y -3 3
Equation of the line passing through point (1, 2) is y - y1 = m2(x - x1), where
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m2 is slope.
or, y - 2 = m (x - 1)........... (ii)
Since line (ii) is parallel to (i), their slopes are equal m1 = m2 = 4
3
Putting the value of m2 is equation (ii)
y - 2 = 4 (x - 1)
3
or, 3y - 6 = 4x - 4
? 4x - 3y + 2 = 0
Therefore the required equation is 4x - 3y + 2 = 0.
Alternative Method
The equation of given straight line is 4x - 3y + 12 = 0 .............(i)
The equation of any straight line parallel to (i) is 4x - 3y + k = 0....... (ii)
The line (ii) passes through the point (1, 2)
4.1 - 3.2 + k = 0
or, 4 - 6 + k = 0
? k=2
Putting the value of k in equation (ii) we get, 4x - 3y + 2 = 0
Therefore, the required equation is 4x - 3y + 2 = 0.
Example 9. Find the equation of straight line perpendicular to the line
Solution: 2x - 3y + 4 = 0 passing through the point (2, 1).
The equation of given straight line is 2x - 3y + 4 = 0 .......... (i)
Its slope (m1) = - coefficient of x
coefficient of y
2
= - 2 = 3
(-3)
The equation a line passing through (2, 1) is y - y1 = m2(x - x1),
where m2 is slope
or, y - 1 = m2 (x - 2) ............... (ii)
Since the line (i) is perpendicular to (ii), we get
m1m2 = -1
or, 2 m2 = - 1
3
3
? m2 = - 2
Putting the value of m1 in equation (ii), we get
y-1=- 3 (x - 2)
2
or, 2y - 2 = - 3x + 6
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or, 3x + 2y = 8
Therefore, the required equation is 3x + 2y = 8.
Alternative Method:
The equation of given line is 2x - 3y + 4 = 0 ........ (i)
Equation of any line perpendicular to (i) is 3x + 2y + k = 0 .......... (ii)
where k is a constant to be determined.
The line (ii) passes through the point (2, 1)
3.2 + 2.1 + k = 0
or, 6 + 2 = - k ? k=-8
Putting the value of k in equation (ii), we get,
3x + 2y - 8 = 0
or, 3x + 2y = 8
Therefore, the required equation is 3x + 2y = 8.
Example 10. Find the equation of a straight line passing through the point (-2, 5) and
perpendicular to the line joining (3, 4) and (2, 8).
Solution: The equation of a line passing through the point (-2, 5) is
y - 5 = m1 (x + 2)......... (i)
where m1 is the slope of line (i)
slope of line joining the points (3, 4) and (2, 8) is m2 = 8-4 = 4 =-4
2-3 -1
Since the two lines are perpendicular to each other,
m1m2 = - 1
or, m1 (-4) = -1
or, m1 = 1
4
Putting the value of m1 in equation (i) , we get
y-5= 1 (x + 2)
4
or, 4y - 20 = x + 2
or, x - 4y + 22 = 0
? x - 4y + 22 = 0
Therefore, the required equation is x - 4y + 22 = 0.
Example 11. Find the equation of the straight line parallel to the line 2x - 3y + 4 = 0 and
passing through the mid-point of the line joining two points (3, 5) and (7, 3)
Solution: Let given line be PQ whose equation is 2x - 3y + 4 = 0 .......... (i)
Let the given points be A(3,5) and B(7, 3).
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( (Mid - point of AB = M 2x – 3y + 4 = 0
x1 + x2 , y1 + y2 P A(3, 5) Q
2 2 S
( (= M
3+7 , 5+ 3 R
2 2 = M (5, 4)
B (7, 3)
Equation of any line parallel to (i) is 2x - 3y + k = 0 ........ (ii)
The line (ii) passes through the point M(5, 4).
So. putting the point M(5, 4) in equation (ii),
we get, 2.5 - 3.4 + k = 0
or, 10 - 12 + k = 0 or, -2 + k = 0
or, k = 2
Put the value of k in equation (ii), we get,
? 2x - 3y + 2 = 0 which is the required equation.
Example 12. Find the equation of the perpendicular bisector of the line joining two points
(5, 7) and (4, 5).
Solution: Let A and B be given two points with coordinates (5, 7) and (4,5) respectively.
Let D be the mid points of AB.
( ( ( (Then, the coordinates of D are 5+4 , 7+ 5 i.e. 9 ,6
2 2 2
Let CD be the perpendicular bisector of AB. C
Slope of AB (m1) = y2 - y1 = 5-7 = -2 = 2
x2 - x1 4-5 -1
Let slope of CD be m2
Then m1.m2 = -1 A D (4, 5) B
or, 2.m2 = –1 (5, 7)
or, m2 = – 1
2
Now, equation of CD is y – y1 = m (x – x1)
or, y – 6 = – 1 x – 9
2 2
2x – 9
or, 2y – 12 = – 2 or, 4y – 24 = –2x + 9
? 2x + 4y = 33 which is the required equation.
Example 13. Find the equation of the straight lines passing through the point (3, -2) and
making an angle of 45° with 6x - 7y - 5 =0
Solution: The equation of straight line passing through point (3, -2) is given by
y - y1 = m(x - x1)
or, y + 2 = m1 (x - 3) ....... (i)
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Its slope is (m) = m1
Again, the equation of given line is 6x - 7y - 5 = 0 ....... (ii) P(3, -2)
Its slope is (m2) = - coefficient of x = - 6 = 6 45° 45°
coefficient of y -7 7 A 6x-7y-5=0B
Angle between the lines (i) and (ii) is 45°
tanT = ± (m1 - m2)
1 + m1m2
m1 - 6
7
or, tan45° = ±
1 + m1 6
or, 7
or, (7m1 - 6) 7
1= ± 7 × 7 + 6m1
7 + 6m1 = ± (7m1 - 6)
Taking positive sign, we get
7 + 6m1 = 7m1, -6 ? m1 = 13
or, -m1 = -13
Taking negative sign, we get
7 + 6m1, = - 7m1 + 6
or, 13m1 = -1 ? m1 = -1
13
Putting the rules of m1 in equation (i),
Putting the values of m1 in equation (i),
Now, for m1 = -1
13
y – y1 = m1(x - x1)
or, y + 2 = -1 (x – 3)
13
or, 13y + 26 = –x + 3
or, x + 13y + 23 = 0
For m1 = 13, equation of the straight line is y + 2 = 13 (x – 3)
or, 13x – y = 41.
Hence, the required equation of lines are x + 13y + 23 = 0 and
13x – y = 41.
Exercise 8.1
Very Short Questions
1. (a) Write the formula to find the angle between two lines y = m1x + c1 and
y = m2x + c2
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(b) Write the conditions of parallelism and perpendicularity of the two lines
y = m1x + c1 and y = m2x + c2
(c) If two lines with equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 are parallel,
show that a1b2 = a2b1
(d) If two lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 are perpendicular to each
other, show that a1a2 + b1b2 = 0
Short Questions
2. Find the acute angle between two lines in each of the following pair of straight lines:
(a) y - (2 - 3 ) x = 5 and y = (2 + 3 ) x + 8.
(b) 2x + 4y = 7 and x + 2y = 5
(c) x = 3 y + 2 and 3 x - y = 5
(d) xcosD + ysinD = p and x sinD - y cosD = q
3. Find the obtuse angle between the lines in each of the following pair of straight lines:
(a) y + 3 x + 8 = 0 and y + 20 = 0
(b) 2x + y = 3 and 3x + 2y = -1
(c) y = 3 x + 8 and 3 x + y + 5 = 0
(d) 3x + 2y = 1 and 2x + 3y + 7 = 0.
4. (a) Show that the lines 3x+4y=10 and 6x + 8y + 10 = 0 are parallel to each other.
(b) Show the lines x + y = 2 and 2x + 2y = 3 are parallel to each other.
(c) Show that the line x-y + 2 = 0 and the line jointing the points (4,6) and (10,12)
are parallel to each other.
(d) Show that the line 3x+y+5 = 0 and the line joining the points (7,0) and (6, 3) are
parallel to each other.
5. (a) Show that the lines 3 x + y = 9 and -x + 3 y + 3 = 0 are perpendicular to
each other.
(b) Show that lines 3x-y- 2 = 0 and x + 3y + 5 = 0 are perpendicular to each other.
(c) Show that the lines joining the points (7, -5) and (3, 4) is perpendicular to the line
4x - 9y + 7 = 0
6. Find the value of k when-
(a) The pair of straight lines kx + 3y - 8 = 0 and 3x - 2y + 11 = 0 are parallel.
(b) The pair of straight lines 2x + 5y + 2 = 0 and kx + y + 6 = 0 are parallel.
(c) The pair of lines 3 x + 2y = 9 and kx + 3 y + 3 = 0 are perpendicular to each
other.
(d) The lines represented by equations kx + 3y + 5 = 0 and 4x - 3y + 10 = 0 are
perpendicular to each other.
(e) The line passing through the points (-5, 7) and (1, -2) is perpendicular to the lines
joining the points (1, -3) and (4, k)
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(f) The line joining the points (8,2) and (4,1) is parallel to the line joining the points
(6,4) and (2, k)
7. Find the equations of straight lines which-
(a) ppppaaaasssssssseieensssgttthhhthrrrooorouuuugggghhhhttthhhtheeeeppppooooiiinnninttt t(((-42(24,,,,535-))3)aa)annanddnddpppeaparrerpaarleplllneeeldnltidtocoiuctthluhaelrealrtliointnoetehthehhaaelvivnliiinnenggehsahsllvaooivppnieegng-s43lso23.lpoepe6545..
(b)
(c)
(d)
8. Find the equation the straight line which-
(a) passes through the point (2,3) and parallel to the line x - 2y - 2 = 0
(b) passes through the point (4, 3) and parallel to the line 2x + 3y + 12 = 0
(c) passes through the point (-1, -2) and perpendicular to the line 6x + 7y = 42.
(d) passes through the point (-1, -2) and perpendicular to the line 5x+4y+ 20 = 0
(e) passes through the point of intersection of the straight line 3x + 4y = 7 and
5x - 2y = 3 and perpendicular to the line 2x + 3y = 5
(f) passes through the point of intersection of 2x - 3y + 1 = 0 and x + 2y = 3 and
parallel to the line 4x + 3y = 12
(g) divides the line joining the points P(-1, -4) and Q (7, 1) in the ratio of 3:2 and
perpendicular to it.
(h) passes through the centroid of ∆ABC with vertices A(4,5), B(-4, -5) and C(1,2) and
parallel to 7x + 5y = 35.
9. Find the equation of the altitude AD of 'ABC with vertices A(2, 3), B(-4, 1) and C(2, 0)
drawn from the vertex A.
10. If the angle between the lines (a2 - b2) x - (p + q) y = 0 and (p2 - q2) x + (a + b) y = 0 is
90°, prove that (a - b) (p - q) = 1.
Long Questions
11. (a) From the point P(-2, 4), perpendicular PQ is drawn to the line AB: 7x - 4y + 15 = 0
Find the equation of PQ.
(b) Find the equation of the altitude PQ drawn from the vertex P to QR in ∆PQR with
P(2,3), Q(-4, 1) and R(2,0).
12. Find the equation of the perpendicular bisector to the join of the following two points.
(a) P(-3, 5) and Q(-6, 7) (b) M(5,6) and N(7, 10) (c) R(2, 4) and S(-2, -4)
13. (a) In rhombus PQRS, P(2, 4) and R(8, 10) are the opposite vertices. Find the equation
of diagonal QS.
(b) M(5,1) and P(-3, 3) are two opposite vertices of square MNPQ. Find the equation
of diagonal QN.
(c) In a square PQRS, P(2,3) and R(-6, 5) are the end points of diagonal PR. Find the
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equation of the other diagonal.
14. (a) Determine the equation of the straight lines through (1, -4) that make an angle of
45° with the straight line 2x + 3y + 7 = 0
(b) Find the equations of the straight lines passing through the point (3, 2) and making
angle of 45° with the line x - 2y - 3 = 0.
(c) Find the equation of two lines passing through the point (1, - 4) and making an
angle of 45° with the lines 2x - 7y +5 = 0
(d) Find the equation of the straight line passing through the point (3 - 2) and making
an angle of 60° with the line 3x + y - 1 = 0.
(e) Determine the value of m so that 3x - my - 8 = 0 makes an angle of 45° with the
line 3x + 5y - 17 = 0
15. (a) Find the equation of the sides of an equilateral triangle whose vertex is (1, 2) and
base is y = 0.
(b) Find the equation of the sides of right angled isosceles triangle whose vertex is at
(-2, -3) and whose base is x = 0
(c) Find the equation of the sides of right angled isosceles triangle whose vertex is
(2, 4) and the equation of base is x = 0.
1. (a) tanT = ± m1 - m2 (b) m1 = m2, m1 . m2 = -1
1 + m1 . m2
2. (a) 60° (b) 0° (c) 30° (d) 90°
3. (a) 120° (b) 172.87° (c) 120° (d) 172.87°
(c) -2
6. (a) - 9 (b) 2 (d) 9
2 5 4
(e) -1 (f) 3
7. (a) 2x + 3y = 13 (b) 3x - 4y + 8 = 0 (c) 5x + 4y + 22 = 0
(d) 5x + 6y = 50 8.(a) x - 2y + 4 = 0 (b) 2x + 3y = 17 (c) 7x - 6y = 5
(d) 4x - 5y = 6 (e) 3x - 2y - 1 = 0 (f) 4x + 3y = 7
(g) 40x + 25y = 127 (h) 21x + 15y = 17 9. 6x - y = 9
11. (a) 4x + 7y = 20 (b) 6x - y = 9 12.(a) 6x - 4y + 51 = 0
(b) x + 2y = 22 (c) x + 2y = 0 13.(a) x + y = 12 (b) 4x - y = 2
(c) 4x - y + 12 = 0 14.(a) 5x + y = 1, x - 5y = 21
(b) 3x - y = 7, x + 3y = 9 (c) 9x - 5y = 29, 5x + 9y + 31 = 0
(d) y + 2 = 0, 3 x - y = 2 + 3 3 (e) 12, - 3
4
15. (a) 3 x - y + 2 = 3, 3 x + y = 2 + 3 (b) x - y = 1, x + y + 5 = 0
(c) x + y = 6, x - y + 2 = 0
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8.2 Pair of straight Lines
Introduction :
Let us consider equations of two straight lines:
2x + 3y + 5 = 0 .............. (i)
x + y + 6 = 0 .............. (ii)
Above equations can be multiplied to find a single equation representing them.
(2x + 3y + 5) (x + y + 6) = 0
or, 2x2 + xy + 12x + 3xy + 3y2 + 18y + 5x + 5y + 30 = 0
or, 2x2 + 3y2 + 5xy + 17x + 23y + 30 = 0
or, 2x2 + 5xy + 3y2 + 17x + 23y + 30 = 0 ..... (iii)
Equation (iii) represents a pair of lines representing lines having equation (i) and (ii)
General Equation of second degree.
An equation of the form ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 is called the general equation
of the second degree in x and y.
Example : 2x2 + 3xy + 5y2 + 3x + 4y + 15 = 0
This equation is in the form of ax2 + 2hxy + by2 + 2gx + 2fy + c = 0,
where a = 2, h = 3 , b = 5, g = 3 f = 2, c = 15.
2 2
Let us consider two general equations of straight lines
a1x +b1y + c1 = 0 ........... (i)
and a2x + b2y + c2 = 0 .......... (ii)
Combining above two equations, we get (a1x + b1y + c1) (a2x + b2y + c2) = 0 on simplification,
We write, a1a2 x2 + (a1b2 + a2b1) xy + b1b2y2 + (b1c2 + b2c1) y + (c1a2 + c2a1) x + c1c2 = 0
or, ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 ........... (iii)
Where, a = a1a2, a1b2 + a2b1 =2h, b1b2 = b, c1a2 + c2a1 = 2g, b1c2 + b2c1 = 2f, c1c2 = c
The above equation (iii) is called general equation of second degree in x and y. It represents
a pair of lines if it can be factorized into two linear factors otherwise it represents a curve.
Homogeneous Equations :
If the sum of indices of variables x and y in each term of an equation is same, such type of
an equation is called homogeneous equation in x and y.
Examples :
(a) x2 + 2xy + y2 = 0 is a homogeneous equation of second degree in x and y.
(b) x3 + 3x2y + 4xy2 + y3 = 0 is a homogeneous equation of the third degree in x and y.
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General equation of the second degree that represents a pair of lines through the origin.
The general equation of the lines passing through the origin are
a1x + b1y = 0 ..... (i)
a2x + b2y = 0 ...... (ii)
Both of these equations are linear in x and y pass through the origin are in the form of y = mx.
Combining these equations, we get,
(a1x + b1y) (a2x + b2y) = 0
or, a1a2x2 + (a1b2 + a2b1) xy + b1b2y2 = 0
or, ax2 + 2hxy + by2 = 0 ............. (iii)
where a = a1a2, 2h = a1b2 + a2b1 , b = b1b2 .
The above equation (iii) is called homogenous equation of the second degree in x and y
which always represents a pair of lines through the origin.
Example : x2 + 3xy + 2y2 = 0
This equation is in the from of ax2 + 2hxy + by2 = 0
we can write,
x2 + 2xy + xy + 2y2 = 0
or, x(x + 2y) + (y(x + 2y) = 0
or, (x + 2y) (x + 6) = 0
Here, x + 2y = 0 and x + y = 0 are equation of two line passing through the origin.
To show that the homogeneous equation of degree two ax2 + 2hxy + by2 = 0 always
represents a pair of lines through the origin.
Proof : The homogeneous equation of second degree in x and y is
ax2 + 2hxy + by2 = 0
or, y2 + 2h . y + a = 0
x2 b x b
y 2+ 2h y a
or, x b x + b = 0
This equation is a quadratic in y and hence it has two roots.
x
Let the roots of the equations be m1 and m2.
Then, we write,
m1 = y and m2 = y
x x
or, y = m1x and y = m2x
Both of these equations are linear in x and y pass through the origin and are in the form of
y = mx.
Hence, ax2 + 2hxy + by2 = 0 always represents a pair of lines through the origin.
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Alternative Method
The homogeneous equation of second degree in x and y is ax2 + 2hxy + by2 = 0
It can be written as ax2 + (2hy)x + by2 = 0
This equation is quadratic in x in the form of ax2 + bx + c = 0
x = -2hy ± (2hy)2 - 4.a.by2 = 2y(-h± h2 - ab )
2a 2a
( (?
= - h ± h2 - ab y
a
Taking positive sign and negative sign, we get
ax + (h - h2 - ab ) y = 0 ........... (i)
ax + (h + h2 - ab ) y = 0 .......... (ii)
Each of these equations is linear in x and y. Hence they represents straight lines. Moreover,
they are in the form of Ax + By = 0. Hence the lines (i) and (ii) are equations of the lines
passing through the origin.
Note :
If equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents a pair of lines,
then ax2 + 2hxy + by2 = 0 represents a pair of lines through the origin parallel to above
pair of lines.
If ax2 + 2hxy + by2 = 0 represents a pair of lines y = m1x and y = m2x
Then, (y - m2x) (y - m2x) = 0
y2 - (m1 + m2) xy + m1 m2x2 = 0 ........ (i)
Also, ax2 + 2hxy + by2 = 0
or, y2 + 2h xy + a y2 = 0 ........... (ii)
b b
2h a
The above equations (i) and (ii) are identical if (m1 + m2) =- b , m1m2 = b
To find angle between the two straight lines represented by ax2 + 2hxy + by2 = 0.
Given equation is
ax2 + 2hxy + by2 = 0 .......... (i)
We know that equation (i) represents a pair of lines passing through the origin.
Let the lines represented by equation (i) be Y 0
y = m1x ............. (ii) =
x
y = m2x .............. (iii) m
1
– y – m1x = 0
y
Then, m1m2 = a –2h X0 X’
b , m1 + m2 = b
slope of line (ii) = m1 Y’
slope of line (iii) = m2
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Let T be the angle between the two lines (ii) and (iii), then,
tanT = ± m1 - m2 =± (m1 + m2)2 - 4m1m2
1 + m1m2 1 + m1m2
( (= ± -2h 2 - 4 a 4h2 a
( (= ± 2 b b b2 b
- 4
=±
a a+b
1+ b
b
h2 - ab ± 2 h2 - ab
a + b , T = tan-1 a+b
Condition for perpendicularity Y
The two lines represented by ax2 + 2hxy + by2 = 0
are perpendicular if T = 90°, y = m 2x
tan90° = ± 2 h2 - ab X y = m 1x X'
a+b
a+b O
or, cot90° = ±
2 h2 - ab Y'
or, 0 = a + b
? a+b=0
Therefore, when a + b = 0, the lines are perpendicular each other.
Condition for Coincident
Two lines represented by ax2 + 2hxy + by2 = 0 are coincident if T = 0°,
tan0° = ± 2 h2 - ab
a+b
or, h2 - ab = 0
? h2 = ab
Therefore, when a + b = 0 then the lines are coincident.
Worked Out Examples
Example 1. Find the single equation of the lines with equations:
Solution:
(a) x - 3y = 0 and x + 3y = 0 (b) x + 2y + 3 = 0 and x + y = 0
(a) The equations of the lines are
x - 3y = 0 ..........(i)
x + 3y = 0 ........ (ii)
The single equation of the lines (i) and (ii) is
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(x - 3y) (x + 3y) = 0
? x2 - 9y2 = 0 is the required equation.
(b) The equations of the lines is,
x + 2y + 3 = 0 ....... (i)
x + y = 0 ........... (ii)
The single equation of above the lines is
(x + 2y + 3) (x + y) = 0
or, x (x + 2y + 3) + y(x + 2y + 3) = 0
or, x2 + 2xy + 3x + xy + 2y2 + 3y = 0
? x2 + 3xy + 2y2 + 3x + 3y = 0 is the required equation.
Example 2. Find the separate equations of two lines represented by the equation.
Solution:
(a) 6x2 + 5xy - 6y2 = 0 (b) 2x2 - 5xy - 3y2 + 3x + 19y - 20 = 0
(a) Here, 6x2 + 5xy - 6y2 - 0
or, 6x2 + 9xy - 4xy - 6y2 = 0
or, 3x (2x + 3y) - 2y (2x + 3y) = 0
or, (2x + 3y) (3x - 2y) = 0
Either 2x + 3y = 0 ........... (i)
or, 3x - 2y = 0 ............ (ii)
3x - 2y = 0 and 2x + 3y = 0 are the required equations
(b) Here, 2x2 - 5xy - 3y2 + 3x + 19y - 20 = 0
It can be written 2x2 - x (5y - 3) + 19y - 3y2 - 20 = 0
Which is in the form of ax2 + bx + c = 0, where,
a = 2, b = 3 - 5y, c = 19y - 3y2 - 20
Now, x = - b± b2 -4ac
2a
(5y - 3) ± (3 - 5y)2 - 4.2.(19y - 3y2 - 20)
= 2.2
= (5y - 3) ± 9-30y+25y2 - 152y + 24y2 + 160
4
= (5y - 3) ± 49y2 - 182y + 169
4
= (5y - 3) ± (7y - 13)2
4
or, 4x = (5y - 3) ± (7y - 13)
Taking positive sign, we get,
4x = 5y - 3 + 7y - 13
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or, 4x - 12y + 16 = 0
? x - 3y + 4 = 0
Taking negative sign, we get,
4x = (5y - 3) - 7y + 13
or, 4x + 2y - 10 = 0
? 2x + y - 5 = 0
Therefore, the required separate equations are x - 3y + 4 = 0 and 2x + y - 5 = 0.
Example 3. Find the equation of pair of lines represented by the equation
Solution: x2 - 2xycosecD+ y2 = 0
Given equation is
x2 - 2xy cosecD + y2 = 0
or, x2 - 2xy cosecD + y2 (cosec2D -cot2D) = 0
or, x2- 2xy cosecD + y2cosec2D - y2cot2D =0
or, (x - ycosecD)2 - (y cotD) 2 = 0.
or, (x - y cosecD + y cotD) (x - ycosecD - y cotD) = 0
or, {x - y (cosecD - cotD)} {x-y(cosecD + cotD)} = 0
Hence, the required pair of lines are
x - y (cosecD - cotD) = 0 and x - y (cosecD + cotD) = 0
Example 4. Find the angle between the pair of the lines represented by:
Solution:
(a) x2 + 7xy + 3y2 = 0 (b) x2 + xy - 6y2 + 7x + 31y- 18 = 0
(a) Given equation is 2x2 + 7xy + 3y2 = 0 ......... (i)
Comparing it with ax2 + 2hxy + by2 = 0, we get,
a = 3, h= 7 and b =3
2
Let T be the angle between the lines represented by equation (i).
Then, tanT = ± 2 h2 - ab = ± 2 49 - 2.3 =± 49 -24
a+b 4 2.5
? tanT = ± 1 2 +3
Taking positive sign,
tanT = 1 ? tanT = tan45°
? T = 45°
Taking negative sign,
tanT = - 1 or, tanT = tan135°
? T = 135°
Therefore, the required angles between the lines represented by
equation (i) are 45° and 135°
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(b) Here, x2 + xy - 6y2 + 7x + 31y - 18 = 0
Taking the homogeneous terms only, we get
x2 + xy - 6y2 = 0 ........... (i)
Comparing it with ax2 + 2hxy + by2 = 0, we get
a = 1, h = 1 , b = -6.
2
Let T be the angle between the lines represented by equation (i), we get
=±2 h2 - ab 1 + 6 1+24
=±2 4 4
tanT a+b =2 =±2
=±1 1-6 -5
1
2 .5
Taking positive sign, -5
tanT = 1 = tan45° ? T = 45°
Taking negative sign,
tanT = - 1 = tan 135°
? T = 135°
Therefore, the required angles are 45° ans 135°.
Example 5. Find the angle between the pair of lines represented by
Solution: x2 - 2xy cosecD+ y2 = 0
Given equation is x2 - 2xy cosecD + y2 = 0
Example 6. Comparing it to ax2 + 2hxy + by2 = 0, we get
Solution:
a = 1, h = -cosecD, b- 1
Let T be the angle between the lines represented by (i),
tanT = ±2 h2 - ab = ± 2 cosec2D = ±cotD
a+b 1+1
= cot (± D) = tan(90° ± D)
? T = 90 ± D
Show that the lines represented by 3x2 + 8xy - 3y2 = 0 are perpendicular to
each other.
Equation is 3x2 + 8xy - 3y2 = 0 ........... (i)
comparing it with ax2 + 2hxy + by2 = 0,
we get a = 3, h = 4, b = -3
Now, a + b = 3 - 3 = 0
Since a + b = 0, the lines represented by equation (i) are perpendicular to
each other.
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Alternatively
Here, 3x2 + 8xy - 3y2 = 0
or, 3x2 + 9xy - xy - 3y2 = 0
or, 3x(x + 3y) - y(x + 3y) = 0
or, (x + 3y) (3x - y) = 0
Separate equation of lines are :
x + 3y = 0 ............(i)
3x - y = 0 ............. (ii)
slope of line (i), m1 = - 1
3
slope of line (ii), m2 = .3
m1.m2 = 1 × 3 = - 1
3
? m1.m2 = -1, the lines (i) and (ii) are at right angle each other.
Example 7. Find the value of k when the pair of lines represented by
(k+2) x2+8xy+ 4y2=0 are coincident.
Solution: Given equation is (k + 2) x2 + 8xy + 4y2 = 0
comparing it to ax2 + 2hxy + by2 = 0, we get,
a = k + 2, h = 4, b = 4
condition for coincidence h2 = ab
or, 16 = (k + 2). 4
or, 16 = k + 2
4
or, 4 = k + 2
? k=2
Example 8. Find the single equation passing through the point (1, 1) and parallel to the
Solution: lines represented by the equation x2 - 5xy + 4y2 = 0
Given equation is
x2 - 5xy + 4y2 = 0
or, x2 - 4xy - xy + 4y2 = 0
or, x(x - 4y) - y (x - 4y) = 0
or, (x - 4y) (x - y) = 0
Separate equations of lines are
x - 4y = 0 ........... (i) and x - y = 0 .............. (ii)
Equation of any line parallel to (i) is x - 4y + k1 = 0 ......... (iii)
Equation of any line parallel to (ii) is x - y + k2 = 0 ......... (iv)
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Above both lines pass through the point (1, 1)
1 - 4 + k1 = 0 or, k1 = 3
and 1 - 1 + k2 = 0 or, k2 = 0
putting the values of k1 and k2 in equations (iii) and (iv), we get,
x - 4y + 3 = 0 x-y =0
required single equation is given by,
(x - 4y + 3) (x - y) = 0
or, x(x - 4y + 3) - y (x - 4y + 3) = 0
or, x2 - 4xy + 3x - 3x - xy + 4y2 - 3y = 0
? x2 - 5xy + 4y2 + 3x - 3y = 0, which is the required single equation.
Example 9. Find the single equation of straight line passing through the origin and
Solution: perpendicular to the lines represented by x2 + 3xy + 2y2 = 0
Given equation is
x2 + 3xy + 2y2 = 0
or, x2 + 2xy + xy + 2y2 = 0
or, x(x+2y) + y(x + 2y) = 0
or, (x + 2y) (x + y) = 0
Separate equations of lines are
x + 2y = 0 and x + y = 0
Equations of the lines perpendicular to x + 2y = 0 and x + y = 0 and passing
through the origin are respectively 2x - y = 0 and x - y = 0
Required single equation is
(2x - y) (x - y) = 0
or, 2x2 - 3xy + y2 = 0
? 2x2 - 3xy + y2 = 0 is the required equation.
Example 10. Prove that two straight lines represented by the equation
(x2 + y2)sin2D = (xcosE - ysinE)2 makes an angle is 2D.
Solution: The given equation of two straight line is
(x2 + y2)sin2D = (xcosE - ysinE)2
or, x2sin2D + y2sin2D = x2cos2E - 2xy sinE.cosE + y2sin2E
or, x2sin2D - x2cos2E + y2sin2D - y2sin2E + 2xy sinE.cosE = 0
or, x2(sin2D - cos2E) + 2xy sinE.cosE + y2(sin2D - sin2E) = 0
By comparing the equation to ax2 + 2hxy + by2 = 0, we get
a = sin2D - cos2E, 2h = sinE.cosE i.e. h = sinE.cosE
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and b = sin2D - sin2E
we know that
tanT = ± 2 h2 - ab = ± 2 (sinE.cosE)2 - (sin2D - cos2E) (sin2D - sin2E)
a+b sin2D - cos2E + sin2D - sin2E
= ± 2sinD (1 - sin2D)
2sin2D - (sin2E + cos2E)
2sinD.cosD sin2D
= ± 2sin2D - 1 = ± cos2D = ± tan2D
or, tanT = tan2D
? T = 2D
Therefore, the angle two lines represented by given equation is 2D. Proved.
Exercise 8.2
Very Short Questions
1. (a) Write the general equation of second degree in x and y.
(b) Define a homogeneous equation.
(c) What is the angle between the pair of lines represented by ax2 + 2hxy + by2 ?
(d) Write the conditions of coincident and perpendicularity of two lines represented
by equation ax2 + 2hxy + by2 = 0
2. Find the single equation represented by the pair of straight lines:
(a) x - y = 0 and x + y = 0 (b) x + 3y = 0 and x - 3y = 0
(c) 4x + 3y = 0 and x + y = 0 (d) 3x + 4y = 0 and 2x - 3y = 0
3. Find the separate equations of straight lines represented by the following equations:
(a) x2 - y2 = 0 (b) 9x2 - 25y2 = 0
(c) x2 - y2 + x - y = 0 (d) x2 + y2 - 2xy + 2x - 2y = 0
4. Show that each pair of lines represented by equations represent a pair perpendicular lines:
(a) 3x2 + 8xy - 3y2 = 0 (b) 5x2 + 24xy - 5y2 = 0
(c) 6x2 - 5xy - 6y2 = 0
5. Show that each pair of lines represented by equations are coincidence:
(a) 16x2 + 16xy + 4y2 = 0 (b) x2 - 6xy + 9y2 = 0
(c) 4x2 - 12xy + 9y2 = 0 (d) 9x2 - 6xy + y2 = 0
Short Questions
6. Determine a single equation representing the following pair of lines.
(a) x + y + 3 = 0 and x - 2y + 3 = 0
(b) 2x + y - 3 = 0 and 2x + y + 3 = 0
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7. Determine the lines represented by each of the following equations:
(a) x2 + 6xy + 9y2 + 4x + 12y - 5 = 0
(b) x2 + 2xy + y2 - 2x - 2y - 15 = 0
(c) 2x2 + 3xy + y2 + 5x + 2y - 3 = 0
(d) 2x2 + xy - 3y2 + 10y - 8 = 0
8. Prove that two lines represented by the following equations are perpendicular to each
other:
(a) x2 - y2 + x - y = 0 (b) 9x2 - 13xy - 9y2 + 2x - 3y + 7 = 0
9. Find the value of O when the lines represented by each of the following equations are
coincident:
(a) 4x2 + 2Oxy + 9y2 = 0 (b) (O + 2) x2 + 3xy + 4y2 = 0
10. Find the value of O when the following equations represents a pair of line perpendicular
to each other:
(a) Ox2 + xy - 3y2 = 0 (b) (O - 25) x2 + 5xy = 0
(c) (3O + 4) x2 - 48xy - O2y2 = 0
11. Find the angle between the following pair of lines:
(a) x2 + 9xy + 14y2 = 0 (b) 2x2 + 7xy + 3y2 = 0
(c) x2 - 7xy + y2 = 0
(d) 9x2 - 13xy - 9y2 + 2x - 3y + 7 = 0 (e) x2 + 6xy + 9y2 + 4x + 12y - 5 = 0
12. Find the angle between the following pair of lines:
(a) x2 - 2xy cotD - y2 = 0 (b) x2 + 2xy cosecD + y2 = 0
(c) y2 + 2xy cotD - x2 = 0 (d) x2 - 2xy cot2D - y2 = 0
13. Find the two separate equation of two lines represented by the following equations.
(a) x2 + 2xy cotD - y2 = 0 (b) x2 - 2xy secD + y2 = 0
(c) x2 + 2xy cosecD + y2 = 0 (d) x2 - 2xy cot2D – y2 = 0
Long Questions
14. (a) Find the equation of two lines represented by 2x2 + 7xy + 3y2 = 0. Find the point
of intersection of the lines. Also, find the angle between them.
(b) Find the separate equations of two lines represented by x2 - 5xy + 4y2 = 0. Find
the angle between the lines and the points of intersection of the lines.
(c) Find the separate equation of two lines represented by the equation
x2 - 2xy cosecD + y2 = 0. Also find the angle between them.
15. (a) Find the pair of lines parallel to the lines x2 - 3xy + 2y2 = 0 and passing through
the origin.
(b) Find the pair of lines parallel to the lines 2x2 - 7xy + 5y2 = 0 and passing through
the point (1, 2).
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16. (a) Find the pair of lines perpendicular to the lines x2 - 5xy + 4y2 = 0 and passing
through the origin.
(b) Find the equation of two straight lines which pass through the point (2,3) and
perpendicular to the lines x2 - 6xy + 8y2 = 0
17. (a) Find the two separate equations when lines represented by kx2 + 8xy - 3y2 = 0 are
perpendicular to each other.
(b) Find the two separate equations when the lines represented by 6x2 + 5xy - ky2 = 0 are
perpendicular to each other.
18. (a) Show that the pair of lines 3x2 - 2xy - y2 = 0 are parallel to the lines
3x2 - 2xy - y2 - 5x + y + 2 = 0
(b) Show that the pair of lines, 4x2 - 9y2 = 0 and 9x2 - 4y2 = 0 are perpendicular to
each other.
Project Work
19. Write short notes with examples on the following
(a) linear equations (b) quadratic equations (c) homogeneous equations
2. (a) x2 - y2 = 0 (b) x2 - 9y2 = 0 (c) 4x2 + 7xy + 3y2 = 0
(d) 6x2 - xy - 12y2 = 0 3.(a) x + y = 0, x - y = 0
(b) 3x + 5y = 0, 3x - 5y = 0 (c) (x + y + 1) = 0, x - y = 0
(d) x - y = 0, x - y + 2 = 0 6.(a) x2 - xy - 2y2 + 6x - 3y + 9 = 0
(b) 4x2 + 4xy + y2 - 9 = 0 7.(a) x + 3y + 5 = 0, x + 3y - 1 = 0
(b) x + y - 5 = 0, x + y + 3 = 0 (c) 2x + y = 1, x + y + 3 = 0
(d) x - y + 2 = 0, 2x + 3y = 4 9.(a) ± 6 (b) - 23
16
10. (a) 3 (b) 25 (c) -1, 4
11. (a) T = tan–1(±13) (b) 45°, 135° (c) tan-1 ± 35 (d) 90° (e) 0°
12. (a) 90° (b) 90° ± D 2 (d) 90°
(c) 90°
13. (a) x + y (cosecD + cotD) = 0, x - y (cosecD - cotD) = 0
(b) x + y(secD + tanD) = 0, x + (secD - tanD) = 0
(c) x + y(cotD + cosecD) = 0, x - y (cotD - cosecD) = 0
(d) x + y (cosec2D - cot2D) = 0, x - y (cosec2D + cot2D) = 0
14. (a) x + 3y = 0, 2x + y = 0, (0, 0), 45°, 135°
(b) x - 4y = 0, x - y = 0, tan-1 ± 3 , (0, 0)
5
(c) x - y(cosecD + cotD) = 0, x + y(cotD - cosecD) = 0, 90° ± D
15. (a) x - 2y = 0, x - y = 0 (b) x - y + 1 = 0, 2x - 5y + 8 = 0
16. (a) 4x + y = 0, x + y = 0 (b) 4x + y = 11, 2x + y = 7
17. (a) 3x - y = 0, x + 3y = 0 (b) (3x – 2y) = 0, 2x + 3y = 0
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8.3 Conic Section
Conic sections are special types of locus which can be obtained geometrically as the intersection
of a plane and a right circular cone. The following are basic types of conic sections.
(a) Parabola (b) Ellipse (c) Circle (d) Hyperbola
Conic sections are also called conics. They play vital roles in the field of mathematics,
engineering, aeronautics etc. In present days, conics are very important in the study of basic
level mathematics.
The path of missile in the form of projectile and path of cables of suspension bridges are
parabolic in nature. In present age of science and technology, parabolic metal surfaces are
used in television dishes, telescopes, headlights, etc.
The planets have elliptical paths while moving around the sun. A smooth elliptical surface
can be used to reflect sound waves and light rays from one focus to another.
Definition : A conic section is the intersection of a plane and a right circular cone.
Before defining different types of conic section, let us define a right circular cone.
Conic means "cone" and section means "to cut" or "to divide".
8.3 Cone
A cone has a closed curved surface as its base Upper A axis AG
and the lateral surface is generated by a line Nappie G T
segment that rotates around the perimeter of Lower Semi vertical V
the closed curved surface keeping one end Nappie
fixed as the vertex. If the base is circle then D angle
the cone is a circular cone. Generator
From above figure we have the following V Vertex
terms :
Semi-vertical angle
The angle D is called semi-vertical angle.
If we rotate the line VG about the line AV so that the angle D does not change.
The surface generated is double napped right circular hollow cone.
Vertex
The point V is called the vertex.
Axis
The fixed line AV is called the axis of the cone.
Generator
The rotating line VG is called the generator of the cone. It is also slant height of the cone.
Nappies
The vertex divides the cone into two parts called nappies.
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8.5 Conic section as a plane section of a right circular cone
On intersecting a right circular cone by a plane in different positions, different sections so
obtained are called conic sections or conics. The shape of conic section depends on the
position of the intersecting plane with respect to cone and by the angle made by it with the
axis of the cone.
(a) Circle V Circle
O
The section of a right circular cone cut by a plane Plane
parallel to its base represents a circle. In other words,
when a cutting plane is at right angles to axis of the
cone, then the section is called circle.
(b) Parabola V Plane
The section of a right circular cone by a plane parallel D
to a generator of the cone represents a parabola. In T
other words, when a cutting plane is inclined to the
axis such that it is parallel to the generator of the cone, Parabola
then the section so obtained is called a parabola. If
T is a angle made by the plane with the axis of cone, O
then D = T.
(c) Ellipse D
T
Section of a right circular cone T
cut by a plane not parallel to any
generator and not parallel or not D
perpendicular to the axis of the cone
is called ellipse. In other words, 183
when a cutting plane is inclined at
some angle to the axis of the cone,
then the section so formed is called
a ellipse. In this case D < T < 90°,
where D is the semi-vertical angle
and T is the angle made by the plan
ze with the axis of the cone.
(d) Hyperbola
When the double cone is cut with
a plane parallel to the axis, we get
hyperbola.
In other words, when the cutting
plane is inclined to the axis such
that it also cuts the other part of
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the double cone, the section is called hyperbola with two branches.
In this case T < D, where T is the angle made by the plane with the axis of the cone.
Exercise 8.3
1. (a) Name conic section formed by a plane with right circular cone parallel to its base?
(b) Name conic section formed by a plane with right circular cone parallel to its
generator.
(c) Name conic section formed by plane with right circular cone inclined at some
angle to axis (but not 90°)
2. Name the following geometrical shapes:
(a) (b) (c) (d)
3. Name the curves which is intersection of plane and cone in the following figures:
(a) (b)
V V
(c) V (d)
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Project Work
4. Cutting a potato or cucumber, folding a paper how a parabola or an ellipse or a hyperbola
can be formed. Discuss in the classes.
5. Take a potato or radish or carrot, make a base of right circular cone. Draw an axis of
the cone on a paper. Then cut the slices of potato or radish or carrot perpendicular to
the base, parallel to the base at different angles to axis. Identity the slices so formed
parabola or ellipse or hyperbola or circle.
1. (a) circle (b) parabola (c) ellipse (d) hyperbola
2. (a) circle (b) ellipse (c) Parabola (d) hyperbola
3. (a) parabola (b) circle (d) ellipse
8.6 Circle
Introduction :
Let us draw a circle with centre at the Y
origin and radius 5 units as shown in the
graph. Discuss the answers of the following C (–3, 4)
questions :
(a) Take two points A(5,0) and B(-5, 0) on the O A (5, 0) X
circumference find the mid-point of AB. Y'
(b) Take point C(-3, 4) on the circumference, X' B (–5, 0)
join OC. Find the length of OA, OB
and OC, Are they equal ? Is there any
common name for them ?
(c) Is AB2 = AC2 + BC2 ?
(d) Find the slopes of AC and BC, is there any relations of slopes of AC and BC?
(e) What is equation of locus of points on above circle ?
Discuss the answers of above questions and draw conclusions from the discussion.
Definition of Circle
The locus of a moving point which moves in such a way that its distance from a fixed point
is always constant is said to be a circle. The fixed point is called centre of the circle and the
constant distance is called the radius of the circle.
In the above figure in graph, O(0,0) is the centre, OA, OB or OC are radii of the circle. Then,
we write OA = OB = OC = r, radius of the circle.
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Equations of Circles Y
(a) Equation of a circle with centre at the P(x, y)
origin (standard form) r
O
Let us draw a circle with centre at the X' X
origin. Let us take P(x, y) on the circle.
Then OP = r, is the radius of the circle.
Then by using distance formula.
OP = r
Squaring on both sides, we get
OP2 = r2 Y'
or, (x - 0)2 + (y - 0)2 = 0
? x2 + y2 = r2 which is the required equation.
Example 1. Find the equation of a circle with centre at the origin and radius 6 units.
Solution: Here, radius of the circle, r = 6 units centre = 0(0,0)
Equation of required circle is,
x2 + y2 = r2
or, x2 + y2 = 62
? x2 + y2 = 36
(b) Equation of a circle with centre at other than origin (Central Form)
Let C(h, k) be the centre of circle and radius CP = r. Let Y
P(x, y) be any point on the circle. Then, by using distance
formula, P(x, y)
We get C(h, k)
CP = r X' X
or, CP2 = r2 [squaring on both sides, we get] O
or, (x - h)2 + (y - k)2 = r2
which is the equation of circle called central form. Y'
Example 2. Find the equation of the circle whose centre is at (4, 5) and radius 6 units.
Solution: Here, centre (h, k) = (4, 5)
Radius (r) = 6 units
Equation of required circle is
(x - h)2 + (y - k)2 = r2
or, (x - 4)2 + (y - 5)2 = 62
or, x2 - 8x + 16 + y2 - 10y + 25 = 36
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or, x2 + y2 - 8x - 10y + 5 = 0
The required equation of circle is x2 + y2 - 8x - 10y + 5 = 0
Particular Cases Y C(h, k) X
r=k
(i) Equation of a circle touching the X - axis. X' O
Let a circle with centre at C(h, k) touch the X-axis at A. Y' A
Then radius r = AC = k.
Equation of the circle on given by
(x - h)2 + (y - k)2 = k2
Example 3. Find the equation of a circle with centre (4, 5) and touches the X - axis.
Solution: Here, centre (h, k) = (4, 5) Y
Since the circle touches X- axis.
Then r = k = 5 X' O C(4, 5)
Now, equation of the circle is, Y' r=5
(x - h)2 + (y - k)2 = k2 X
or, (x - 4)2 + (y - 5)2 = 52
or, x2 - 8x + 16 + y2 - 10y + 25 = 25
? x2 + y2 - 8x - 10y + 16 = 0 Y
which is the required equation of the circle.
(ii) Equation of a circle touching the Y-axis. B hC
Let a circle with centre c(h, k) touch the Y-axes at B. k
Then radius r = BC = h. X' O A X
Now, equation of the circle is (x - h)2 + (y - k)2 = h2 Y'
Example 4. Find the equation of the circle with centre (-2, -4) and touching the Y-axis.
Solution:
Here, centre (h, k) = (-2 -4) Y
Since the circle touches Y-axis r = h = -2 X' OX
Now, equation of the circle is,
(x - h)2 + (y - k)2 = h2 (-2, -4)
or, (x + 2) + (y + 4)2 = (-2)2
or, x2 + 4x + 4 + y2 + 8y + 16 = 4 Y'
? x2 + y2 + 4x + 8y + 16 = 0 is the required equation.
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(iii) Equation of a circle touching both of the axes X and Y. Y
Let a circle with centre (h, k) touch
both of the axes X and Y at A and B. B h
Then AC = BC = h = k = r C
Then equation of the circle is
X' k X
(x - h)2 + (y - h)2 = h2 OA
or, (x - k)2 + (y - k)2 = k2.
Y'
Example 5. Find the equation of the circle with centre (3,3) Y
Solution: touching both of the axes.
B 3 C(3, 3)
Here, centre (h, k) = (3, 3) 3
Since the circle touch both of the axis, X' OA X
h=k=r=3 Y'
Now, equation of the circle is
(x - h)2 + (y -h)2 =h2
or, (x - 3)2 + (y - 3)2 = 32
or, x2 - 6x + 9 + y2 - 6y + 9 =9
? x2 + y2 - 6x - 6y + 9 = 0 is the required equation of the circle.
(C) Equation of a circle in diameter form
Let A(x1, y1) and B(x2, y2) be the P(x,y) ends of a diameter P(x, y)
AB of a circle and O is the its centre.
Let P(x,y) be any point on the circle. Join AP and BP A B
Then by plane geometry, (x1, y1) (x2, y2)
APB = 90° [ Angle at semi-circle is right angle.]
Now, slope of AP = (m1) = y - y1
slope of BP = m2 = y -xy-2x2
x- x2
Since AP is perpendicular to PB,
we write, m1.m2 = - 1
or, y - y1 . y - y2 =-1
x- x1 x- x2
or, (y - y1) (y - y2) = - (x - y1) (x - x2)
or, (x - x1) (x - x2) + (y - y1) (y - y2) = 0
which is the required equation circle in diameter form.
The following properties of circles are to be noted
(i) The straight line joining the centre of the circle and middle point of a chord is
perpendicular to the chord.
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(ii) The perpendicular drawn from the centre of the circle on the chord bisects it.
(iii) The angle at semi-circle is a right angle.
(iv) A straight line is tangent to the circle if the length of the perpendicular drawn from the
centre to the line is equal to the radius of the circle.
(v) Two circles touch externally if the distance between the centres of two circles is equal
to the sum of radii of the circles. Two circles touch internally if the distance between
the centres of two circles is equal to difference of their radii.
r1 r2 P C1
C1 C2 C2
Externally touched circles Internally touched circles
r1 + r2 = C1 C2 PC1 = r1, PC2 = r2
r1 – r2 = c1 c2
Example 6. Find the equation of a circle whose ends of diameter are (-2, -3) and (2,3).
Solution: The ends of a diameter are (-2. -3) and (2,3).
i.e. (x1, y1) = (-2, -3), (x2, y2) = (2, 3)
Now, equation of circle is given by,
(x - x1) (x - x2) + (y - y1) (y - y2) = 0
or, (x + 2) (x - 2) + (y + 3) (y - 3) = 0
or, x2 - 4 + y2 - 9 = 0
? x2 + y2 = 13 is the required equation of circle.
(d) General Equation of the circle
Consider the equation
x2 + y2 + 2gx + 2fy + c = 0.......... (i)
In which the coefficients of x2 and y2 are equal each having unity and there is no term
containing xy.
The above equation (i) can be written as
x2 + 2gx + y2 + 2fy = -c
or, x2 + 2gx + g2 + y2 + 2fy + f2 = g2 + f2 - c
or, (x + g)2 + (y + f)2 = ( g2 + f2 - c )2 .......... (ii)
which is in the form of
(x - h)2 + (y - k)2 = r2 ........... (iii)
Hence, the above equation of (i) is equation of circle. Comparing equation (ii) to (iii)
Centre = (h, k) = (-g, -f)
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radius (r) = g2 + f2 - c
The above equation (i) is called general equation of circle.
Worked out Examples
Example 1. Find the equation of circle with centre (4,1) and radius 5.
Solution: Here, centre (h, k) = (4, 1)
Example 2. radius (r) = 5
Solution: Now, equation of the circle is
Example 3. (x -h)2 + (y - k)2 = r2
Solution: or, (x - 4)2 + (y - 1)2 = 52
or, x2 - 8x + 16 + y2 - 2y + 1 = 25
190 ? x2 + y2 - 8x - 2y = 8 is the required equation.
Find the centre and radius of the circle x2 + y2 + 4x - 6y + 4 = 0
The given equation is
x2 + y2 + 4x - 6y + 4 = 0
or, x2 + 4x + 4 + y2 - 6y + 9 = 4 + 9 - 4
or, (x + 2)2 + (y - 3)2 = 32
which is in the form of (x-h)2 + (y - k)2 = r2,
? centre (h, k) = (-2, 3), radius (r) = 3 units.
If the equation of circle 4x2 + 4y2 - 24x - 20y - 28 = 0, find the coordinates of
the centre and diameter of the circle.
Equation of given circle is
4x2 + 4y2 - 24x - 20y - 28 = 0
or, x2 + y2 - 6x - 5y - 7 = 0
or, 5 52 (= 7 + 32 + 5 (2
2 2 2
( (or, 25
x2 - 2.3.x + 32 + y2 - 2.y.( + +
(x - 3)2 + 9+ 4
y- 5 ((2
2 =7
(
(or, y- 5 ( 2 89
(x - 3)2 + 2 (=4
( (or, y- 5 2 89 2
(x - 3)2 + 2 = 2
which is in the form of
(x - h)2 + (y - k)2 = r2
(centre (h, k) = 3, 5 and radius (r) = 89 units.
2 2
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Example 4. Find the equation of a circle whose ends of diameter are (2,4) and (4, 5)
Solution: The ends of diameter are (x1y1) = (2, 4) and (x2,y2) = (4, 5)
Now, equation of circle in diameter form is
(x - x1) (x - x2) + (y - y1) (y - y2) = 0
or, (x - 2)(x - 4) + (y -4)(y - 5) = 0
or, x2 - 6x + 8 + y2 - 9y + 20 = 0
? x2 + y2 - 6x - 9y + 28 = 0 is the required equation.
Example 5. Find the equation of the circle whose centre is the point of intersection of
Solution: x + 2y - 1 = 0 and 2x - y - 7 =0 and which passes through the point (3, 1).
Given equation are
x + 2y - 1 = 0 ...... (i)
2x - y - 7 = 0 ....... (ii)
We get from equation (i), (3, 1) r= 2x–y–7=0
x = 1 - 2y .......... (iii) P
C x+y–1=0
Putting the value of x in equation (ii),
2(1 - 2y) - y - 7 = 0
or, 2 - 4y - y - 7 = 0
y=-1
Putting the value of y in equation (iii), we get,
x = 1 - 2(-1) = 3
? centre of circle is C (3, – 1)
Radius = CP = (3 – 3)2 + (1 + 1)2 = 2 units.
Now, equation of circle is,
(x – h)2 + (y – k)2 = r2
or, (x – 3)2 + (y + 1)2 = 22
or, x2 – 6x + 9 + y2 + 2y + 1 = 4
? x2 + y2 – 6x + 2y + 6 = 0 is the required equation.
Example 6. Find the equation of a circle with centre (6, 5) and touching the x-axis.
Solution:
Here, centre of the circle = (h, k) = (6, 5) Y
Since the circle touches the X-axis C(6, 5)
Radius = r = k = 5 r=5
Now, the equation of circle is X' O X
(x – h)2 + (y – k)2 = k2
or, (x – 6)2 + (y – 5)2 = 52 Y'
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or, x2 – 12x + 36 + y2 – 10y + 25 = 25
Example 7. or, x2 + y2 – 12x – 10x + 36 = 0
Solution: ? x2 + y2 – 12x – 10y + 36 = 0 is the required equation of the circle.
Find the equation of a circle passing through the origin and making
Example 8. intercepts 4 and 6 on positive x and y axes.
Solution:
Let X-intercept OA=4 and Y-intercept OB=6 units of given circle. Let C be the
centre of the circle, draw CM perpendicular to OA and CN perpendicular to
OB. Y
Then,ON = 1 OB = 1 × 6 = 3 B
2 2
1 1
OM = 2 ×OA = 2 ×4 = 2 N C (2, 3)
Then centre of circle has coordinates (2,3) 3
X' 2
Now, radius of circle (r) = OC = 22 + 32 A X
OM
= 4 + 9 = 13
Y'
Equation of required circle is
(x - h)2 + (y - k)2 = r2
or, (x - 2)2 + (y - 3)2 = ( 13 )2
or, x2 - 4x + 4 + y2 - 6y + 9 = 13
or, x2 + y2 - 4x - 6y = 0
? x2 + y2 - 4x - 6y = 0 is the required equation.
Find the equation of a circle passing through the points (1,2), (3,4) and (5,2).
Let, C(h,k) be the centre of circle which passes through the points M(1,2),
N(5,2) and P (3,4) P (3, 4)
Then CM = CN = CP
CM2 = CN2 = CP2 (radii of same circle) M (1, 2) C (h, k) N (5, 2)
CM2 = (1 - h)2 + (2 - k)2
= 1 - 2h + h2 + 4 - 4k + k2
= h2 + k2 - 2h - 4k + 5 ...... (i)
CN2 = (5-h)2 + (2 - k)2
= 25 - 10h + h2 + 4 - 4k + k2
= h2 + k2 - 10h - 4k + 29 ..... (ii)
CP2 = (3 - h)2 + (4 - k)2
= 9 - 6h + h2 + 16 - 8k + k2
= h2 + k2 - 6h - 8k + 25 ...... (iii)
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Taking CM2 = CN2 [from (i) and (ii)]
h2 + k2 - 2h - 4k + 5 = h2 + k2 - 10h - 4k + 29
or, 8h = 24
? h=3
Again, taking CN2 = CP2
or, h2 + k2 - 10h - 4k + 29 = h2 + k2 - 6h - 8k + 25
or, - 4h + 4k = - 4
or, h - k = 1
putting the value of h in this equation, k = 2
centre (h, k) = (3, 2)
and radius (r) = Distance between C(3,2) and M(1,2)
= (1-3)2 + (2-2)2 = 2 units.
Now, equation of the required circle is (x-h)2 + (y-k)2 = r2
or, (x - 3)2 + (y -2)2 = 22
or, x2 - 6x + 9 + y2 - 4y + 4 = 4
? x2 + y2 - 6x - 4y + 9 =0 which is required equation.
Example 9. Find the equation of circle touching coordinate axes at (6,0) and (0,6)
Solution:
The circle touches the X-axis at A(6,0). Y-axis at B(0,6). Draw CA and CB
perpendicular on X and Y-axis respectively. Then OACB will be a square.
OA = BC = 6 Y
OB = AC = 6
centre of the circle is C(6,6)
and radius r = 6. B (0, 6) C (6, 6)
Now, equation of the circle is given by
(x - h)2 + (y - k)2 = r2 X' O X
or, (x - 6)2 + (y - 6)2 = 62 A (6, 0)
or, x2 - 12x + 36 + y2 - 12y + 36 = 36 Y'
? x2 + y2 - 12x - 12y + 36 = 0 is the required equation.
Example 10. Find the equation of a circle which passes through the points (6,5) and (4,1)
and with centre on the line 4x + y = 16
Solution: Let C(h, k)be the centre of the circle. The centre (h,k) lies on the line
4x + y = 16 4h + k = 16 ...... (i)
Also, let equation of circle be (x-h)2 + (y-k)2 = r2 ....... (ii)
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The circle (ii) passes through the points
P(6,5) and Q (4,1) Q (4, 1) P (6, 5)
or, (h - 6)2 + (k - 5)2 = (h - 4)2 + (k + 1)2
or, -4h - 8k = -44
or, h + 2k = 11.........(iii) C (h, k)
solving equations (i) and (iii), we get
h = 3 and k = 4
centre = C (3,4)
Radius (r) = CP = (6-3)2 + (5-4)2 = 32 + 12 = 10 units
Now, equation of the circle is
(x -h)2 + (y - k)2 = r2
or, (x-3)2 + (y - 4)2 = 10
or, x2 - 6x + 9 + y2 - 8y + 16 = 10
? x2 + y2 - 6x - 8y + 15 = 0 is the required equation.
Example 11. A line y - x = 2 cuts a circle x2 + y2 + 2x = 0 at two distinct points. Find the
coordinates of the points. Also, find the length of the chord.
Solution: Given equation of the line is y - x = 2 ......... (i)
The line (i) cuts the circle at two distinct points A and B.
We solve equations (i) and (ii) to find the coordinates of A and B.
From equation (i), y = x + 2
Put the value of y in equation (ii), we get
x2 + (x+2)2 + 2x = 0 x2 + y2 + 2x = 0
or, x2 + x2 + 4x + 4 + 2x = 0
or, 2x2 + 6x + 4 = 0
or, x2 + 3x + 2 = 0
or, x2 + 2x + x + 2 =0 y–x=2 B
or, x(x + 2) + 1(x + 2) = 0 A
or, (x + 2) (x + 1) = 0
x = -1, -2
when x = -1, y = -1 + 2 = 1
when x = -2, y = -2 + 2 = 0
The coordinates of A and B are (-1, 1) and (-2, 0) respectively.
Now, the length of chord AB is given by
AB = (-2+1)2+(0 -1)2 = 1 + 1= 2 units.
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Example 12. Find the equation of a circle which touches the X-axis at the point (6,0) and
passes through the points (3,3).
Solution: Let C(h,k) be the centre of the circle. Since it touches X-axis at D(6,0).
radius (r) = k and h = 6. Y
Now, the equation of circle is
(x - h)2 + (y - k)2 = k2
or, (x - h)2 + (y - k)2 = k2
It passes through the point (6,6), we get C = (h, k)
(3 - 6)2 + (3 - k)2 = k2
or, 9 + 9 - 6k + k2 = k2 (3, 3)
? k=3 X' O D (6, 0) X
Therefore, the required equation is Y'
(x - 6)2 + (y - 3)2 = 32
or, x2 - 12x + 36 + y2 - 6y + 9 = 9
? x2 + y2 - 12x - 6y + 36 = 0 is the required equation of circle.
Example 13. Show that two circles x2 + y2 = 100 and x2 + y2 - 24x - 10y + 160 = 0 touch
each other externally.
Solution: The equation of given circles are
x2 + y2 = 100.......... (i)
x2 + y2 - 24x - 10y + 160 = 0 .......... (ii)
From equation (i), radius (r1) = 10 units r1 = 13 r2 = 3
centre C1 = (0, 0) C1 = (0, 0) C2 = (12, 5)
From equation (ii),
radius = g2 + f2 - c
(r2) = (-12)2 + (-5)2 - 160
= 144 + 25 - 160
= 9 = 3 units
Centre (C2) = (12, 5)
Distance between the centres = (12 - 0)2 + (5 - 0)2
= 144 + 25
= 169 = 13 units
Sum of the radii of the circle = 10 + 3 = 13
Since the distance between the centres of circles is equal to sum of the radii
of circles, two circles touch externally.
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Example 14. Find the equation of the circle which touches the positive Y-axis at a distance of
4 units from the origin and cuts of an intercept 6 from the X-axis.
Solution : Let C (h, k) be the centre of the circle touching Y-axis at N and ON = 4.
From C, draw perpendiculars CM and CN on X-axis and Y-axis respectively.
Then, ON = CM = 4 =k
k=4 Y
Also, AM = MB = 1 AB = 3,
2
From right angled triangle CMA,
AC2 = AM2 + CM2 N C (h, k)
= 9 + 16 = 25 4
X' O A 4 BX
AC = radius (r) = 5 units 3M
(h, k) = (5, 4)
Now, equation of the circle is Y'
(x - h)2 + (y - k)2 = r2
or, (x - 5)2 + (y - 4)2 = 52
or, x2 - 10x + 25 + y2 - 8y + 16 = 25
or, x2 + y2 - 10x - 8y + 16 = 0
? x2 + y2 - 10x - 8y + 16 = 0 is the required equation.
Exercise 8.4
Very Short Questions
1. (a) Write down the equation of a circle with centre at O(o,o) and radius R.
(b) Write down the equation of a circle with centre (h,k) and radius r.
(c) Write radius and centre of general circle x2 + y2 + 2gx + 2fy + c = 0
(d) Write the equation of a circle whose ends of a diameter are (x1, y1) and (x2, y2) .
2. Write the equation of the circle under the following conditions.
(a) centre at O(0,0) and radium 5 units.
(b) centre at (-2, -3) and radius 6 units.
(c) ends of a diameter are (1,2) and (3,4).
(d) ends of a diameter are (4,5) and (-2, -3).
(e) ends of a diameter are (-a,0) and (a,0).
3. Write the centre and radius of the following circles.
(a) (x - 2)2 + (y - 4)2 = 25 (b) (x + 3)2 + (y + 5)2 = 81
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(c) (x - 2)2 + (y + 7)2 = 36 (d) (x - a)2 + (y - b)2 = c2
4. Find the centre and radius of the following circles :
(a) x2 + y2 = 81 (b) x2 + y2 + 6x + 4y - 12 = 0
(c) x2 + y2 + 2px - 2qy + p2 + q2 = m2 + n2
(d) 4x2 + 4y2 + 16x - 24y - 52 = 0
(e) 9x2 + 9y2 - 36x + 6y = 107 (f) x2 + y2 - 2axcosT -2aysinT = 0
Short Questions
5. (a) Find the equation a circle whose centre is the (4,5) and touches x-axis.
(b) Find the equation of a circle with circle with centre (-3, 5) and touching x-axis.
(c) Centre (1,3) and touching the y-axis.
6. Find the equations of circles under the following condition.
(a) Centre at (4, -1) and through (-2, -3)
(b) radius 5 units and touching positive x-axis and y-axis.
(c) touching the coordinate axes at (a,0) and (a,0).
Long Questions
7. (a) Find the equation of a circle whose centre is the point of intersection of
x + 2y - 1 = 0 and 2x - y - 7 = 0 and passing through the point (6,4)
(b) Find the equation of a circle with equations of two diameters are x + y = 14 and
2x - y = 4 and passing through the point (4, 5).
8. Determine the points of intersections of a straight line and circle in each of the following
cases:
(a) x + y = 3, x2 + y2 - 2x - 3 = 0 (b) 2x - y + 1 = 0, x2 + y2 = 2
(c) x + y = 2, x2 + y2 = 4
Also find the lengths of the intercepts (chords) in each case.
9. Find the centre and radius of circles.
(a) passing through the points P(-4, -2), Q(2,6) and R(2, -2)
(b) passing through the points P(2, -1), Q(2, 3) and R(4, 1)
10. Find the equations of circle.
(a) passing through the points (5,7),(6,6) and (2, -2).
(b) passing through the points (-6,5), (-3,-4) and (2,1)
11. (a) Find the equation of circle passing through the origin and the point (4,2) and
centre lies on the lines x + y = 1.
(b) Find the equation of the circle passing through the points (3,2) and (5,4) and
centre lies on the line 3x - 2y = 1.
12. (a) Find the equation of the circle which touches the x-axis at (4,0) and cuts of an
intercept of 6 units from the y-axis positively.
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(b) Find the equation of the circle which touches the y-axis at (0, 4) and cuts of an
intercept 6 units from the positive part of x-axis.
13. (a) Prove that the points A(2, -4), B(3, -1), C(3, -3) and D(0,0) are concyclic.
(b) Prove that the points P(3,3), Q(7,1), R(4,0) and S(6,4) are concyclic.
14. (a) Show that the two circles x2 + y2 = 36 and x2 + y2 - 12x - 16y + 84 = 0 touch
externally.
(b) Show that the two circles x2 + y2 = 81 and x2 + y2 - 6x - 8y + 9 = 0 touch
internally.
Project Work
15. Draw a circle of radius 5 units taking centre at (2,3). On a graph, find equation of
the circle. Take any four points on the circle and show that the points satisfy the
equation of the circle. Discuss the conclusion.
2. (a) x2 + y2 = 25 (b) x2 + y2 + 4x + 6y = 23 (c) x2 + y2 - 4x - 6y + 11 = 0
(d) x2 + y2 - 2x - 2y - 23 = 0 (e) x2 + y2 = a2
3. (a) (2, 4), 5 (b) (-3, -5), 9 (c) (2, -7), 6 (d) (a, b), c
4. (a) (0, 0), 9 (b) (-3, -2), 5 (c) (-p, q), m2 + n2
(d) (-2, 3), 26 (e) 2, - 1 , 4 (f) (acosT, asinT), a
3 (b) x2 + y2 + 6x - 10y + 9 = 0
5. (a) x2 + y2 - 8x - 10y + 16 = 0
(c) x2 + y2 - 2x - 6y + 9 = 0 6.(a) x2 + y2 - 8x + 2y = 23
(b) x2 + y2 - 10x - 10y + 25 = 0 (c) x2 + y2 - 2ax - 2ay + a2 = 0
7. (a) x2 + y2 - 6x + 2y = 24 (b) x2 + y2 - 12x - 16y + 87 = 0
8. (a) (1, 2), (3, 0), 2 2 (b) (-1, -1), 1 , 7 , 65 (c) (0, 2), (2, 0), 2 2
5 5 2
9. (a) (-1, 2), 5 (b) (2, 1), 2
10. (a) x2 + y2 - 4x - 6y - 12 = 0 (b) x2 + y2 + 6x - 2y = 15
11. (a) x2 + y2 - 8x + 6y = 0 (b) x2 + y2 - 6x - 8y + 21 = 0
12. (a) x2 + y2 - 8x - 10y + 16 = 0 (b) x2 + y2 - 10x - 8y + 16 = 0
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vedanta Excel In Opt. Mathematics - Book 10 9
Trigonometry
9.0 Review of Trigonometricy Ratios of Compound Angle
Let A and B be any two angles, the sum (A+ B) and the difference (A - B) are known as
compound angles. The following are the list of formula of compound angles.
(i) sin (A + B) = sinA.cosB + cosA. sinB
(ii) cos (A + B) = cosA. cosB - sinA. sinB
(iii) tan (A + B) = tanA + tanB
1 - tanA tanB
cotA.cotB-1
(iv) cot (A + B) = cotB + cotA
(v) sin (A -B) = sinA. cosB - cosA. sinB
(vi) cos (A - B) = cosA.cosB + cosA. sinB
(vii) tan (A -B) = tanA - tanB
1 + tanA.tanB
cotA.cotB + 1
(viii)cot (A-B) = cotB - cotA
(ix) sin (A + B). sin (A - B) = sin2A - sin2B = cos2B - cos2A
(x) cos(A + B).cos (A - B) = cos2A - sin2B = cos2B - sin2A
(xi) tan(A + B). tan(A - B) = tan2A – tan2B
1 - tan2A.tan2B
cot2A.cot2B-1
(xii) cot(A + B). cot (A - B) = cot2B - cot2A
9.1 Trigonometric Ratios of Multiple Angles
Let A be any angle. Then, 2A, 3A, 4A, ... etc. are called multiple angles of A. In this
sub-units, we discuss the trigonometric ratios of angles 2A and 3A.
I. Trigonometric Ratios of 2A
We know that sin (A + B) = sinA.cosB - cosA.sinB.
(a) sin2A = 2sinA.cosA = 1 2tanA = 1 2cotA
+ tan2A + cot2A
Here, sin2A = sin(A + A) = sinA.cosA + cosA.sinA
? sin2A = 2sinA.cosA............ (i)
sin2A 2sinA.cosA = 2sinA.cosA
1 sin2A + cos2A
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2sinA.cosA
cos2A
= sinA +cosA = 2tanA
tan2A + 1
cos2A
2tanA
? sin2A = 1 + tan2A ...............(ii)
2tanA 2 1 1 2cot2A 2cotA
1 + tan2A cotA cotA cot2A + 1 + cot2A
Also, sin2A = = = . =
1 1
1 + cot2A
2cotA
? sin2A = 1 + cot2A ......... (iii)
Combining above (i), (ii) and (iii), we get,
sin2A = 2sinA.cosA = 1 2tanA = 1 2cotA
+ tan2A + cot2A
1-tan2A cot2A -1
(b) cos2A = cos2A - sin2A = 1 - 2sin2A = 2cos2A - 1 = 1 + tan2A = 1 + cot2A
Here, cos(A + B) = cosA.cosB - sinA.sinB
Now, cos2A = cos(A + A) = cosA.cosA - sinA.sinA
? cos2A = cos2A - sin2A ............ (i)
cos2A = cos2A - sin2A = 1 - sin2A - sin2A
? cos2A = 1 - 2sin2A ............... (ii)
or, 2sin2A = 1 - cos2A.
? cos2A = 2cos2A - 1 .............(iii)
2cos2A = 1 + cos2A
Again, cos2A = cos2A - sin2A
cos2A-sin2A
cos2A
= cos2A - sin2A = sin2A + cos2A = 1- tan2A
sin2A + cos2A 1 + tan2A
cos2A
1-tan2A
? cos2A = 1+tan2A ........... (iv)
1 - 1 cot2A - 1
cot2A cot2A + 1
and cos2A = = .......... (v)
1
1 + cot2A
Combining (i), (ii), (iii), (iv), and (v), we get
cos2A = cos2A - sin2A = 1 - 2sin2A = 2cos2A - 1 = 1- tan2A = cot2A -1
1 + tan2A cot2A + 1
2tanA
(c) tan2A = 1 - tan2A
Here, tan(A + A) = tanA + tanA = 2tanA
1 - tanA.tanA 1 - tan2A
2 tanA
? tan2A = 1 – tan2A
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