NCERT Solutions Class 11th Mathematics. FREE Flip-BOOK by Study Innovations. 614 Pages

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Q7 :
Find the sum to n terms of the series 3 x 8 + 6 x 11 + 9 x 14 +…
Answer :
The given series is 3 × 8 + 6 × 11 + 9 × 14 + …
an = (nth term of 3, 6, 9 …) × (nth term of 8, 11, 14, …)
= (3n) (3n + 5)
= 9n2 + 15n

Q8 :
Find the sum to n terms of the series 12 + (12 + 22) + (12 + 22 + 32) + …

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Answer :
The given series is 12 + (12 + 22) + (12 + 22 + 32 ) + …
an = (12 + 22 + 32 +…….+ n2)

= n(n+1)(2n+1)6=n(2n2+3n+1)6=2n3+3n2+n6=13n3+12n2+16n

Q9 :
Find the sum to n terms of the series whose nth term is given by n (n + 1) (n + 4).
Answer :
an = n (n + 1) (n + 4) = n(n2 + 5n + 4) = n3 + 5n2 + 4n

Q10 :
Find the sum to n terms of the series whose nth terms is given by n2 + 2n
Answer :
an= n2 + 2n

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Consider
The above series 2, 22, 23, … is a G.P. with both the first term and common ratio equal to 2.
Therefore, from (1) and (2), we obtain

Q11 :
Find the sum to n terms of the series whose nth terms is given by (2n - 1)2
Answer :
an = (2n – 1)2 = 4n2 – 4n + 1

Exercise Miscellaneous : Solutions of Questions on Page Number : 199
Q1 :

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Show that the sum of (m + n)th and (m - n)th terms of an A.P. is equal to twice the mth term.

Answer :
Let a and d be the first term and the common difference of the A.P. respectively.
It is known that the kth term of an A. P. is given by
ak = a + (k -1) d
∴ am + n = a + (m + n -1) d
am - n = a + (m - n -1) d
am = a + (m -1) d
∴ am + n + am - n = a + (m + n -1) d + a + (m - n -1) d
= 2a + (m + n -1 + m - n -1) d
= 2a + (2m - 2) d
= 2a + 2 (m - 1) d
=2 [a + (m - 1) d]
= 2am
Thus, the sum of (m + n)th and (m - n)th terms of an A.P. is equal to twice the mth term.

Q2 :
If the sum of three numbers in A.P., is 24 and their product is 440, find the numbers.

Answer :
Let the three numbers in A.P. be a - d, a, and a + d.
According to the given information,
(a - d) + (a) + (a + d) = 24 … (1)
⇒ 3a = 24
∴a=8
(a - d) a (a + d) = 440 … (2)
⇒ (8 - d) (8) (8 + d) = 440
⇒ (8 - d) (8 + d) = 55
⇒ 64 - d2 = 55
⇒ d2 = 64 - 55 = 9
⇒d=±3
Therefore, when d = 3, the numbers are 5, 8, and 11 and when d = -3, the numbers are 11, 8, and 5.
Thus, the three numbers are 5, 8, and 11.

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Q3 :
Let the sum of n, 2n, 3n terms of an A.P. be S1, S2 and S3, respectively, show that S3 = 3 (S2- S1)
Answer :
Let a and b be the first term and the common difference of the A.P. respectively.
Therefore,

From (1) and (2), we obtain

Hence, the given result is proved.
Q4 :
Find the sum of all numbers between 200 and 400 which are divisible by 7.
Answer :
The numbers lying between 200 and 400, which are divisible by 7, are
203, 210, 217, … 399
∴First term, a = 203
Last term, l = 399

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Common difference, d = 7
Let the number of terms of the A.P. be n.
∴ an = 399 = a + (n –1) d
⇒ 399 = 203 + (n –1) 7
⇒ 7 (n –1) = 196
⇒ n –1 = 28
⇒ n = 29

Thus, the required sum is 8729.

Q5 :
Find the sum of integers from 1 to 100 that are divisible by 2 or 5.
Answer :
The integers from 1 to 100, which are divisible by 2, are 2, 4, 6… 100.
This forms an A.P. with both the first term and common difference equal to 2.
⇒100 = 2 + (n –1) 2
⇒ n = 50

The integers from 1 to 100, which are divisible by 5, are 5, 10… 100.
This forms an A.P. with both the first term and common difference equal to 5.
∴100 = 5 + (n –1) 5
⇒ 5n = 100
⇒ n = 20

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The integers, which are divisible by both 2 and 5, are 10, 20, … 100.
This also forms an A.P. with both the first term and common difference equal to 10.
∴100 = 10 + (n –1) (10)
⇒ 100 = 10n
⇒ n = 10

∴Required sum = 2550 + 1050 – 550 = 3050
Thus, the sum of the integers from 1 to 100, which are divisible by 2 or 5, is 3050.

Q6 :
Find the sum of all two digit numbers which when divided by 4, yields 1 as remainder.
Answer :
The two-digit numbers, which when divided by 4, yield 1 as remainder, are
13, 17, … 97.
This series forms an A.P. with first term 13 and common difference 4.
Let n be the number of terms of the A.P.
It is known that the nth term of an A.P. is given by, an = a + (n –1) d
∴97 = 13 + (n –1) (4)
⇒ 4 (n –1) = 84
⇒ n – 1 = 21
⇒ n = 22
Sum of n terms of an A.P. is given by,

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Thus, from (1) and (2), we obtain
Q25 :
Find the sum of the following series up to n terms:
Answer :
The nth term of the given series is

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Q26 :
Show that
Answer :
nth term of the numerator = n(n + 1)2 = n3 + 2n2 + n
nth term of the denominator = n2(n + 1) = n3 + n2

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From (1), (2), and (3), we obtain

Thus, the given result is proved.

Q27 :

A farmer buys a used tractor for Rs 12000. He pays Rs 6000 cash and agrees to pay the balance in annual
installments of Rs 500 plus 12% interest on the unpaid amount. How much will be the tractor cost him?

Answer :
It is given that the farmer pays Rs 6000 in cash.
Therefore, unpaid amount = Rs 12000 – Rs 6000 = Rs 6000
According to the given condition, the interest paid annually is
12% of 6000, 12% of 5500, 12% of 5000, …, 12% of 500
Thus, total interest to be paid = 12% of 6000 + 12% of 5500 + 12% of 5000 + … + 12% of 500
= 12% of (6000 + 5500 + 5000 + … + 500)
= 12% of (500 + 1000 + 1500 + … + 6000)
Now, the series 500, 1000, 1500 … 6000 is an A.P. with both the first term and common difference equal to 500.
Let the number of terms of the A.P. be n.
∴ 6000 = 500 + (n – 1) 500
⇒ 1 + (n – 1) = 12
⇒ n = 12

∴Sum of the A.P
Thus, total interest to be paid = 12% of (500 + 1000 + 1500 + … + 6000)
= 12% of 39000 = Rs 4680
Thus, cost of tractor = (Rs 12000 + Rs 4680) = Rs 16680

Q28 :

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Shamshad Ali buys a scooter for Rs 22000. He pays Rs 4000 cash and agrees to pay the balance in annual
installment of Rs 1000 plus 10% interest on the unpaid amount. How much will the scooter cost him?

Answer :
It is given that Shamshad Ali buys a scooter for Rs 22000 and pays Rs 4000 in cash.
∴Unpaid amount = Rs 22000 – Rs 4000 = Rs 18000
According to the given condition, the interest paid annually is
10% of 18000, 10% of 17000, 10% of 16000 … 10% of 1000
Thus, total interest to be paid = 10% of 18000 + 10% of 17000 + 10% of 16000 + … + 10% of 1000
= 10% of (18000 + 17000 + 16000 + … + 1000)
= 10% of (1000 + 2000 + 3000 + … + 18000)
Here, 1000, 2000, 3000 … 18000 forms an A.P. with first term and common difference both equal to 1000.
Let the number of terms be n.
∴ 18000 = 1000 + (n – 1) (1000)
⇒ n = 18

∴ Total interest paid = 10% of (18000 + 17000 + 16000 + … + 1000)
= 10% of Rs 171000 = Rs 17100
∴Cost of scooter = Rs 22000 + Rs 17100 = Rs 39100

Q29 :
A person writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four
different persons with instruction that they move the chain similarly. Assuming that the chain is not broken
and that it costs 50 paise to mail one letter. Find the amount spent on the postage when 8th set of letter is
mailed.

Answer :
The numbers of letters mailed forms a G.P.: 4, 42, … 48
First term = 4
Common ratio = 4
Number of terms = 8
It is known that the sum of n terms of a G.P. is given by

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It is given that the cost to mail one letter is 50 paisa.

∴Cost of mailing 87380 letters = Rs 43690

Thus, the amount spent when 8th set of letter is mailed is Rs 43690.

Q30 :
A man deposited Rs 10000 in a bank at the rate of 5% simple interest annually. Find the amount in 15th year
since he deposited the amount and also calculate the total amount after 20 years.

Answer :
It is given that the man deposited Rs 10000 in a bank at the rate of 5% simple interest annually.

∴ Interest in first year

∴Amount in 15th year = Rs
= Rs 10000 + 14 × Rs 500
= Rs 10000 + Rs 7000
= Rs 17000

Amount after 20 years =
= Rs 10000 + 20 × Rs 500
= Rs 10000 + Rs 10000
= Rs 20000

Q31 :

A manufacturer reckons that the value of a machine, which costs him Rs 15625, will depreciate each year by
20%. Find the estimated value at the end of 5 years.

Answer :
Cost of machine = Rs 15625

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Machine depreciates by 20% every year.
Therefore, its value after every year is 80% of the original cost i.e., of the original cost.

∴ Value at the end of 5 years = = 5 × 1024 = 5120

Thus, the value of the machine at the end of 5 years is Rs 5120.

Q32 :

150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on second day,
4 more workers dropped out on third day and so on. It took 8 more days to finish the work. Find the number
of days in which the work was completed.

Answer :
Let x be the number of days in which 150 workers finish the work.
According to the given information,
150x = 150 + 146 + 142 + …. (x + 8) terms
The series 150 + 146 + 142 + …. (x + 8) terms is an A.P. with first term 146, common difference –4 and number of
terms as (x + 8)

However, x cannot be negative.
∴x = 17

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Therefore, originally, the number of days in which the work was completed is 17.
Thus, required number of days = (17 + 8) = 25

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NCERT Solutions for Class 11 Maths Chapter 10

Straight Lines Class 11

Chapter 10 Straight Lines Exercise 10.1, 10.2, 10.3, miscellaneous, miscellaneousmiscellaneous Solutions
Exercise 10.1 : Solutions of Questions on Page Number : 211
Q1 :
Draw a quadrilateral in the Cartesian plane, whose vertices are (-4, 5), (0, 7), (5, -5) and (-4, -2). Also, find its
area.
Answer :
Let ABCD be the given quadrilateral with vertices A (–4, 5), B (0, 7), C (5, –5), and D (–4, –2).
Then, by plotting A, B, C, and D on the Cartesian plane and joining AB, BC, CD, and DA, the given quadrilateral can
be drawn as

To find the area of quadrilateral ABCD, we draw one diagonal, say AC.
Accordingly, area (ABCD) = area (ΔABC) + area (ΔACD)
We know that the area of a triangle whose vertices are (x1, y1), (x2, y2), and (x3, y3) is
Therefore, area of ΔABC

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Area of ΔACD

Thus, area (ABCD)
Q2 :
The base of an equilateral triangle with side 2a lies along they y-axis such that the mid point of the base is at
the origin. Find vertices of the triangle.
Answer :
Let ABC be the given equilateral triangle with side 2a.
Accordingly, AB = BC = CA = 2a
Assume that base BC lies along the y-axis such that the mid-point of BC is at the origin.
i.e., BO = OC = a, where O is the origin.
Now, it is clear that the coordinates of point C are (0, a), while the coordinates of point B are (0, –a).
It is known that the line joining a vertex of an equilateral triangle with the mid-point of its opposite side is
perpendicular.

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Hence, vertex A lies on the y-axis.

On applying Pythagoras theorem to ΔAOC, we obtain or (0, a), (0, –a), and
(AC)2 = (OA)2 + (OC)2
⇒ (2a)2 = (OA)2 + a2
⇒ 4a2 – a2 = (OA)2
⇒ (OA)2 = 3a2
⇒ OA =

∴Coordinates of point A =

Thus, the vertices of the given equilateral triangle are (0, a), (0, –a), and

.

Q3 :

Find the distance between and when: (i) PQ is parallel to the y-axis, (ii) PQ is parallel
to thex-axis.

Answer :

The given points are and .

(i) When PQ is parallel to the y-axis, x1 = x2.

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In this case, distance between P and Q

(ii) When PQ is parallel to the x-axis, y1 = y2.
In this case, distance between P and Q

Q4 :
Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).

Answer :
Let (a, 0) be the point on the x axis that is equidistant from the points (7, 6) and (3, 4).

On squaring both sides, we obtain
a2 – 14a + 85 = a2 – 6a + 25
⇒ –14a + 6a = 25 – 85
⇒ –8a = –60

Thus, the required point on the x-axis is .

Q5 :
Find the slope of a line, which passes through the origin, and the mid-point of
the line segment joining the points P (0, -4) and B (8, 0).

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Answer :
The coordinates of the mid-point of the line segment joining the points

P (0, –4) and B (8, 0) are
It is known that the slope (m) of a non-vertical line passing through the points (x1, y1) and (x2, y2) is given by

.
Therefore, the slope of the line passing through (0, 0) and (4, –2) is

.

Hence, the required slope of the line is .

Q6 :
Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and (-1, -1) are the vertices of a right
angled triangle.
Answer :
The vertices of the given triangle are A (4, 4), B (3, 5), and C (–1, –1).
It is known that the slope (m) of a non-vertical line passing through the points (x1, y1) and (x2, y2) is given by

.

∴Slope of AB (m1)

Slope of BC (m2)

Slope of CA (m3)
It is observed that m1m3 = –1
This shows that line segments AB and CA are perpendicular to each other
i.e., the given triangle is right-angled at A (4, 4).
Thus, the points (4, 4), (3, 5), and (–1, –1) are the vertices of a right-angled triangle.

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Q7 :
Find the slope of the line, which makes an angle of 30° with the positive direction of y-axis measured
anticlockwise.
Answer :
If a line makes an angle of 30° with the positive direction of the y-axis measured anticlockwise, then the angle made
by the line with the positive direction of the x-axis measured anticlockwise is 90° + 30° = 120°.

Thus, the slope of the given line is tan 120° = tan (180° – 60°) = –tan 60°
Q8 :
Find the value of x for which the points (x, -1), (2, 1) and (4, 5) are collinear.
Answer :
If points A (x, –1), B (2, 1), and C (4, 5) are collinear, then
Slope of AB = Slope of BC

Thus, the required value of x is 1.

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Q9 :
Without using distance formula, show that points (-2, -1), (4, 0), (3, 3) and
(-3, 2) are vertices of a parallelogram.
Answer :
Let points (–2, –1), (4, 0), (3, 3), and (–3, 2) be respectively denoted by A, B, C, and D.

Slope of AB

Slope of CD =
⇒ Slope of AB = Slope of CD
⇒ AB and CD are parallel to each other.

Now, slope of BC =

Slope of AD =
⇒ Slope of BC = Slope of AD
⇒ BC and AD are parallel to each other.
Therefore, both pairs of opposite sides of quadrilateral ABCD are parallel. Hence, ABCD is a parallelogram.
Thus, points (–2, –1), (4, 0), (3, 3), and (–3, 2) are the vertices of a parallelogram.

Q10 :
Find the angle between the x-axis and the line joining the points (3, -1) and (4, -2).
Answer :

The slope of the line joining the points (3, –1) and (4, –2) is
Now, the inclination (θ) of the line joining the points (3, –1) and (4, – 2) is given by

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tan θ= –1
⇒ θ = (90°+ 45°) = 135°
Thus, the angle between the x-axis and the line joining the points (3, –1) and (4, –2) is 135°.

Q11 :

The slope of a line is double of the slope of another line. If tangent of the angle between them is , find the
slopes of he lines.

Answer : be the slopes of the two given lines such that .
Let

We know that if θisthe angle between the lines l1and l2with slopes m1and m2, then .
It is given that the tangent of the angle between the two lines is .

Case I

If m = –1, then the slopes of the lines are –1 and –2.

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If m = , then the slopes of the lines are and –1.
Case II

If m = 1, then the slopes of the lines are 1 and 2.

If m = , then the slopes of the lines are .

Hence, the slopes of the lines are –1 and –2 or and –1 or 1 and 2 or .

Q12 : . If slope of the line is m, show that .
A line passes through
Answer :

The slope of the line passing through is .
It is given that the slope of the line is m.

Hence,
Q13 :

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If three point (h, 0), (a, b) and (0, k) lie on a line, show that .

Answer :
If the points A (h, 0), B (a, b), and C (0, k) lie on a line, then
Slope of AB = Slope of BC

On dividing both sides by kh, we obtain

Hence,
Q14 :
Consider the given population and year graph. Find the slope of the line AB and using it, find what will be the
population in the year 2010?

Answer :

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Since line AB passes through points A (1985, 92) and B (1995, 97), its slope is
Let y be the population in the year 2010. Then, according to the given graph, line AB must pass through point C
(2010, y).
∴Slope of AB = Slope of BC

Thus, the slope of line AB is , while in the year 2010, the population will be 104.5 crores.

Exercise 10.2 : Solutions of Questions on Page Number : 219
Q1 :
Write the equations for the x and y-axes.

Answer :
The y-coordinate of every point on the x-axis is 0.
Therefore, the equation of the x-axis is y = 0.
The x-coordinate of every point on the y-axis is 0.
Therefore, the equation of the y-axis is x = 0.

Q2 :

Find the equation of the line which passes through the point (–4, 3) with slope .

Answer : , whose slope is m, is .
We know that the equation of the line passing through point

Thus, the equation of the line passing through point (–4, 3), whose slope is , is

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Q3 :
Find the equation of the line which passes though (0, 0) with slope m.

Answer :

We know that the equation of the line passing through point , whose slope is m, is .

Thus, the equation of the line passing through point (0, 0), whose slope is m,is
(y – 0) = m(x – 0)

i.e., y = mx

Q4 :

Find the equation of the line which passes though and is inclined with the x-axis at an angle of
75°.

Answer :
The slope of the line that inclines with the x-axis at an angle of 75° is
m = tan 75°

We know that the equation of the line passing through point , whose slope is m, is .

Thus, if a line passes though and inclines with the x-axis at an angle of 75°, then the equation of the line
is given as

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Q5 :

Find the equation of the line which intersects the x-axis at a distance of 3 units to the left of origin with slope
-2.

Answer :
It is known that if a line with slope m makes x-intercept d, then the equation of the line is given as
y = m(x - d)
For the line intersecting the x-axis at a distance of 3 units to the left of the origin, d = -3.
The slope of the line is given as m = -2
Thus, the required equation of the given line is
y = -2 [x - (-3)]
y = -2x - 6
i.e., 2x + y + 6 = 0

Q6 :
Find the equation of the line which intersects the y-axis at a distance of 2 units above the origin and makes
an angle of 30° with the positive direction of the x-axis.

Answer :
It is known that if a line with slope m makes y-intercept c, then the equation of the line is given as
y = mx + c

Here, c = 2 and m = tan 30° .

Thus, the required equation of the given line is

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Q7 : .
Find the equation of the line which passes through the points (-1, 1) and (2, -4).

Answer :

It is known that the equation of the line passing through points (x1, y1) and (x2, y2) is
Therefore, the equation of the line passing through the points (–1, 1) and
(2, –4) is

Q8 :
Find the equation of the line which is at a perpendicular distance of 5 units from the origin and the angle
made by the perpendicular with the positive x-axis is 30°

Answer :
If p is the length of the normal from the origin to a line and ̉ۡis the angle made by the normal with the positive
direction of the x-axis, then the equation of the line is given by xcos ̉ۡ + y sin ̉ۡ= p.
Here, p = 5 units and É = 30°
Thus, the required equation of the given line is
x cos 30° + y sin 30° = 5

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Q9 :
The vertices of ΔPQR are P (2, 1), Q (-2, 3) and R (4, 5). Find equation of the median through the vertex R.
Answer :
It is given that the vertices of ΔPQR are P (2, 1), Q (–2, 3), and R (4, 5).
Let RL be the median through vertex R.
Accordingly, L is the mid-point of PQ.
By mid-point formula, the coordinates of point L are given by

It is known that the equation of the line passing through points (x1, y1) and (x2, y2) is .
Therefore, the equation of RL can be determined by substituting (x1, y1) = (4, 5) and (x2, y2) = (0, 2).

Hence,

Thus, the required equation of the median through vertex R is .

Q10 :

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Find the equation of the line passing through (-3, 5) and perpendicular to the line through the points (2, 5)
and (-3, 6).

Answer :

The slope of the line joining the points (2, 5) and (–3, 6) is
We know that two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals
of each other.

Therefore, slope of the line perpendicular to the line through the points (2, 5) and (–3, 6)
Now, the equation of the line passing through point (–3, 5), whose slope is 5, is

Q11 :

A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in the ratio 1:n. Find the
equation of the line.

Answer :
According to the section formula, the coordinates of the point that divides the line segment joining the points (1, 0)
and (2, 3) in the ratio 1: n is given by

The slope of the line joining the points (1, 0) and (2, 3) is

We know that two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals
of each other.

Therefore, slope of the line that is perpendicular to the line joining the points (1, 0) and (2, 3)

Now, the equation of the line passing through and whose slope is is given by

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Q12 :
Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point
(2, 3).
Answer :
The equation of a line in the intercept form is

Here, a and b are the intercepts on x and y axes respectively.
It is given that the line cuts off equal intercepts on both the axes. This means that a = b.
Accordingly, equation (i) reduces to

Since the given line passes through point (2, 3), equation (ii) reduces to
2+3=a⇒a=5
On substituting the value of a in equation (ii), we obtain
x + y = 5, which is the required equation of the line

Q13 :
Find equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is
9.
Answer :
The equation of a line in the intercept form is

Here, a and b are the intercepts on x and y axes respectively.

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It is given thata + b = 9 ⇒ b = 9 – a … (ii)
From equations (i) and (ii), we obtain
It is given that the line passes through point (2, 2). Therefore, equation (iii) reduces to

If a = 6 and b = 9 – 6 = 3, then the equation of the line is
If a = 3 and b = 9 – 3 = 6, then the equation of the line is

Q14 :
Find equation of the line through the point (0, 2) making an angle with the positive x-axis. Also, find the
equation of line parallel to it and crossing the y-axis at a distance of 2 units below the origin.
Answer :
The slope of the line making an angle with the positive x-axis is

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Now, the equation of the line passing through point (0, 2) and having a slope is .

The slope of line parallel to line is .

It is given that the line parallel to line crosses the y-axis 2 units below the origin i.e., it passes
through point (0, –2).

Hence, the equation of the line passing through point (0, –2) and having a slope is

Q15 :
The perpendicular from the origin to a line meets it at the point (- 2, 9), find the equation of the line.
Answer :
The slope of the line joining the origin (0, 0) and point (–2, 9) is
Accordingly, the slope of the line perpendicular to the line joining the origin and point (– 2, 9) is

Now, the equation of the line passing through point (–2, 9) and having a slope m2 is

Q16 :
The length L (in centimetre) of a copper rod is a linear function of its Celsius temperature C. In an
experiment, if L = 124.942 when C = 20 and L = 125.134 when C = 110, express L in terms of C.

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