Homogeneous equation of second degree

Homogeneous equation of second degree
If the sum of the powers of x and y in each term of an equation is equal then such equation is
called homogeneous equation in x and y.
For example:
(i) 3x – 5y = 0 is the homogeneous equation of first degree.
2
2
(ii) 3x – 5xy + 2y = 0 is the homogeneous equation of second degree.
3
2
2
3
(iii) x + 5x y – 4xy – y = 0 is the homogeneous equation of third degree.
Let us consider the two linear equations passing through the origin.
a1x + b1y = 0 …………(i)
a2x + b2y = 0 ……………..(ii)

Now multiplying equations (i) and (ii), we get
(a1x + b1y) (a2x + b2y) = 0
2
2
or, a1a2x + a1b2xy + a2b1xy + b1b2y = 0
2
2
or, a1a2x + (a1b2 + a2b1) xy + b1b2y = 0

putting a1a2 = a, a1b2 + a2b1 = 2h and b1b2 = b
we get
2
2
ax + 2hxy + by = 0 …(iii)

Here, in equation(iii), the power of each term of the variable is 2. This equation is called the
homogeneous equation of second degree in x and y. It represents a pair of straight lines
passing through origin.

2
2
Note: If g = f = c = 0, then general equation ax + 2hxy + by + 2gx + 2fy + c = 0 is reduced
2
2
to ax + 2hxy + by = 0 is known as homogeneous equation of second degree in x and y.

Homogeneous equation of the second degree in x and y represents a pair of straight lines
passing through origin.

Proof: Let the homogeneous equation of second degree in x and y be
2
2
ax + 2hxy + by = 0 ……………(i)
ax 2 2hxy by 2 o
or, + + =
a a a a
2h b
2
2
or, x + xy + y = 0
a a
hy  2  2 b
hy
hy
2
2
or, x + 2.x. +   –   + y = 0
a     a
a
a
2 2
 hy 2 h y by 2
or, x +  = 2 –
 a  a a
2 2
 hy 2 h y – aby 2
or, x +  = 2
 a  a
2
2
 hy  y (h – ab)
or, x +  = 
 a  a 2
hy y
2
or, x + =  h – ab
a a
–hy y
2
or, x =  h – ab
a a
RAMHARI SHRESTHA [dhadinge@hotmail.com]

2
(– h  h – ab)
or, x = . y
a
Taking +ve sign, we get
2
(– h + h – ab)y
x = …(ii)
a

Taking –ve sign, we get
2
(– h – h – ab)y
x = … (iii)
a

Here equation (ii) and equation (iii) are the first degree in x and y and independent of the
2
constant term and hence represent the pair of lines through the origin. Hence, ax +2hxy+by²
=0 always represents a pair of lines through the origin.

2
2
Angle between a pair of lines represented by the equation ax + 2hxy + by = 0.

The homogeneous equation of second degree in x and y is
2
2
ax + 2hxy + by = 0
2
2
or, by + 2hxy + ax = 0

Now, dividing both sides by b, we get
2h a
2
2
y + xy + x = 0……………(i)
b b

Since, the homogeneous equation of second degree in x and y always represents a pair of
lines passing through origin, let them be
y = m1x
or, y – m1x = 0………………(ii)
and y = m2x
or, y – m2x = 0……………(iii)

Now, multiplying eq(ii) and eq(iii), we get
(y– m1x) (y – m2x) = 0
2
2
or, y – m1xy – m2xy + m1m2x = 0
2
2
or, y – (m1 + m2) xy + m1m2x = 0 ………………(iv)

Now, comparing equation (i) and (iv), we get
– 2h a
m1 + m2 = , m1 .m2 =
b b

2
Now, m1 – m2 = (m1 + m2) – 4m1 . m2


 – 2h 2 a
=   – 4 
 b  b
4h 2 4a
–
= 2
b b
2
4h – 4ab
=
b 2


RAMHARI SHRESTHA [dhadinge@hotmail.com]

2
2
= h – ab
b
Let be the angle between two lines
y = m1x and y = m2x

m1 – m2
Then, tan = 
1 + m1.m2
2 2
b (h – ab)
or, tan =  a
1 +
b
2 2
b (h – ab)
or, tan =  b + a

b
2
2 h – ab
or, tan = 
a + b
2
 2 h – ab 
–1
  = tan  
 a + b 

2
2
which is the angle between the pair of lines represented by ax + 2hxy + by = 0.
2
2
Similarly the angle between the lines represented by ax + 2hxy + by + 2gx + 2fy + c = 0
2
 2 h – ab 
is also,  = tan  a + b  .
–1



Condition for perpendicularity
2
2
When the lines represented by ax + 2hxy + by = 0 are perpendicular, then º
2
2 h – ab
So, tan90º = 
a + b
2
2 h – ab
 = 
a + b
 a + b = 0

Condition for coincident
2
2
When the lines represented by ax + 2hxy + by = 0 are Coincident, then,  = 0º
2
2 h – ab
So, tan0º  = 
a + b
2
2 h – ab
or, 0 = 
a + b
2
or, 2 h – ab = 0
2
or, h – ab = 0
2
 h – ab = 0
i.e. h² = ab








RAMHARI SHRESTHA [dhadinge@hotmail.com]

Worked out Examples

1. Find the single equation represented by the equations x – 3y + 1 = 0 and 5x + 2y = 0
Solution:
Here, the given equations are
x – 3y + 1 = 0 …(i)
and 5x + 2y = 0…(ii)

Now, multiplying equation (i) and (ii), we get
(x – 3y + 1) (5x + 2y) = 0
or, 5x (x – 3y + 1) + 2y (x – 3y + 1) = 0
2
2
or, 5x – 15xy + 5x + 2xy – 6y + 2y = 0
2
2
or, 5x – 13xy – 6y + 5x + 2y = 0 is the required single equation.

2. Find the separate equations of the lines represented by the following equations.
2
2
(a) 6x – 5xy – 4y = 0
2
2
(b) x + 3xy + 2y + 3x + 5y + 2 = 0
2
2
(c) x + 2xy sec + y = 0.
Solution:
2
2
(a) 6x – 5xy – 4y = 0
2
2
or, 6x – (8 – 3) xy – 4y = 0
2
2
or, 6x – 8xy + 3xy – 4y = 0
or, 2x (3x – 4y) + y(3x – 4y) = 0
or, (2x + y) (3x – 4y) = 0

Either 2x + y = 0…(i)
or, 3x – 4y = 0 …(ii)

Hence equation (i) and (ii) are the required separate equations.
2
2
(b) x + 3xy + 2y + 3x + 5y + 2 = 0
2
2
or, x + 3xy + 3x + 2y + 5y + 2 = 0
2
2
or, x + (3y + 3) x + (2y + 5y + 2) = 0

2
Now comparing with ax + bx + c = 0, we get
2
a = 1, b = 3y + 3, c = 2y + 5y + 2

2
– b  b – 4ac
Now, x =
2a
2
2
– (3y + 3)  (3y + 3) – 4  1 (2y + 5y + 2)
or, x =
2  1
2
2
or, 2x = – 3y – 3  9y + 18y + 9 – 8y – 20y – 8
2
or, 2x + 3y + 3 =  y – 2y + 1
2
or, 2x + 3y + 3 =  (y – 1)
or, 2x + 3y + 3 =  (y – 1 )
Now, taking +ve sign, we get
2x + 3y + 3 = y – 1
or, 2x + 2y + 4 = 0
RAMHARI SHRESTHA [dhadinge@hotmail.com]

 x + y + 2 = 0 …(i)

Taking –ve sign, we get
2x + 3y + 3 = – y + 1
or, 2x + 4y + 2 = 0
 x + 2y + 1 = 0 …(ii)
Hence, equations (i) and (ii) are the required separate equations.

2
2
(c) x + 2xy sec + y = 0.
2
2
2
2
or, x + 2.x. ysec + (ysec) – (ysec) + y = 0

2
2
2
or, (x + ysec) – y sec  + y = 0
2
2
2
or, (x + ysec) – y (sec  – 1) = 0
2
2
2
or, (x + ysec) – y tan  = 0
2
2
or, (x + ysec) – (ytan) = 0
or, (x + ysec – ytan) (x + ysec + ytan) = 0

Either, x + ysec – ytan = 0  x + ( sec  – tan) y = 0 ........ (i)
or, x + ysec + ytan = 0  x + (sec  + tan y = 0 ........ (ii)
Hence, equations (i) and (ii) are the required equations.

2
2
3. Find the angle between the pair of lines represented by 2x – 7xy + 3y = 0
Solution:
2
2
Here, the given equation is 2x – 7xy + 3y = 0

2
2
Now, comparing it with ax + 2hxy + by = 0, we get
– 7
a = 2, 2h = – 7, h = , b = 3
2
2
 2 h – ab
Now, angle between pair of lines is given by tan =
a + b
 7  2
–
 2   – 2× 3

 2
or, tan =
2 + 3
49
–
 2 4 6
or, tan =
5
25
 2 4
or, tan =
5
2 × 5
or, tan = 
2 × 5
or, tan =  1

Now, taking +ve sign, we get
tan = 1
or, tan = tan 45°
  = 45°

Again, taking – ve sign, we get
RAMHARI SHRESTHA [dhadinge@hotmail.com]

tan = – 1
or, tan = – tan 45°
or, tan = tan(180° – 45°)
  = 135º
Hence,  = 45°, 135º

2
2
4. If  is the acute angle between a pair of lines represented by x + 2xy sec + y = 0,
then show that  = 
Solution:
Here, the given equation is
2
2
x + 2xy sec + y = 0
2
2
Now, comparing it with ax + 2hxy + by = 0, we get
a = 1, 2h = 2sec, h = sec, b = 1
Now, angle between pair of lines is given by
2
2 h –ab
tan = 
a+ b
2
 2 sec  – 1× 1
or, tan =
1 + 1
2
 2 sec  – 1
or, tan =
2
or, tan =  tan
since  is acute angle, so taking +ve sign only, we get
tan = tan
 

2
2
5. Show that the lines represented by 4x – 4xy + y = 0 are coincident.
Solution:
2
2
Here, the given equation of a pair of lines is 4x – 4xy + y = 0
2
2
Now, comparing it with ax + 2hxy + by = 0, we get
a = 4, 2h = – 4, h = – 2, b = 1

2
Now, h – ab
2
= (–2) – 4 × 1
= 4 – 4
= 0

2
2
2
Since h – ab = 0, the two lines represented by 4x – 4xy + y = 0 are coincident.

6. Find the value of k when the pair of lines represented by (k + 5) x – 5xy – (3k – 1) y = 0
2
2
are perpendicular.
Solution:
2
2
Here, the given equation of a pair of lines is (k + 5) x – 5xy – (3k – 1) y = 0

2
2
Now, comparing it with ax + 2hxy + by = 0, we get,
–5
a = k + 5, 2h = – 5, h = , b = – (3k – 1)
2



RAMHARI SHRESTHA [dhadinge@hotmail.com]

Since the pair of lines represented by the given equation are perpendicular then
a + b = 0
or, (k + 5) + { – (3k – 1)} = 0
or, k + 5 –3k + 1 = 0
or, – 2k = – 6
 k = 3

7. Find the single equation of a pair of lines passing through (2, – 1) and
2
2
perpendicular to the pair of lines 2x + 7xy + 3y = 0.
Solution:
2
2
Here, the given equation of a pair of lines is 2x + 7xy + 3y = 0
2
2
or, 2x + 6xy + xy + 3y = 0
or, 2x (x + 3y) + y(x + 3y) = 0
or, (x + 3y) (2x + y) = 0

Either, x + 3y = 0 ………… (i)
OR, 2x + y = 0 ……….. (ii)

Now, the equation of a straight line perpendicular to x + 3y = 0 is 3x – y + k = 0
Since it passes through (2, – 1), then
3 × 2 – (–1) + k = 0
 k = –7
Hence, the required equation is 3x – y – 7 = 0 …………. (iii)

Again, the equation of a straight line perpendicular to 2x + y = 0 is x – 2y + k = 0
Since, it passes through (2, – 1), then
2 – 2(–1) + k = 0
 k = – 4

Hence the required equations is
x – 2y – 4 = 0 …………… (iv)

Now, multiplying equation (iii) and (iv), we get
(3x – y – 7) (x – 2y – 4) = 0
2
2
 3x – 7xy + 2y – 19x + 18y + 28 = 0 is the required single equation

Exercise 21

1. Find the single equation representing the following pair of lines.
(a) 5x – 2y = 0 and 3x + 4y = 0
(b) 2x – 5y + 1 = 0 and 5x = 3y.
(c) 7x + 2y = 7 and 4x – 5y + 1 = 0
(d) mx + ny = 0 and nx = my
(e) y + (cot + cosec) x = 0 and y + (cot – cosec)x = 0
(f) y + xsec + xtan = 0 and y + xsec – xtan = 0

2. Find the separate equation of the straight lines represented by the following single equations.




RAMHARI SHRESTHA [dhadinge@hotmail.com]

2
2
2
2
(a) 3x + 5xy + 2y = 0 (b) 4x – 5xy – 6y = 0
2
2
2
(c) 4x – 9y = 0 (d) 5x – 25xy = 0
2
2
2
(e) ax + bxy = 0 (f) x – 6xy + 9y – x + 3y = 0
2
2
2
2
(g) x – y – 5x – 5y = 0 (h) x + 2xy + y – 3x – 3y + 2 = 0
2
2
2
2
(i) x + 3xy + 2y + 3x + 5y + 2 = 0 (j) 2x – 7xy + 3y + 11x – 13y + 12 = 0
2
2
(k) xy – 5x – 4y + 20 = 0 (l) x + 2xy sec + y = 0
2
2
2
2
(m) x + 2xy cosec + y = 0 (n) x + 2xy tan – y = 0
2
2
(o) x – 2xy cot – y = 0 (p) x(x – 1) – y (y – 1) = 0
2
2
2
2
2
2
(q) ab(x + y ) – xy(a + b ) = 0 (r) x – y – 6y – 9 = 0
2
2
2
2
2
2
(s) x + 2xy + y – 2x – 2y – 15 = 0 (t) y cos  – 2xy + (1 + sin ) x = 0
2
2
(u) 4x + 3xy + 2y = 0

3. Find the angle between the pair of lines represented by the following equations.
2
2
2
2
(a) 2x + 7xy + 3y = 0 (b) 3x – 4xy + 3y = 0
2
2
2
2
(c) 2x + xy – y – x + 5y – 6 = 0 (d) 4x + 12xy + 9y = 0
2
2
2
(e) x – y = 0 (f) x + xy = 0
2
2
2
2
(g) 3x – 8xy – 3y = 0 (h) 4x + 4xy – 3y = 0
2
2
2
2
(i) x + 2xy cosec + y = 0 (j) y + 2xy secβ + x = 0
2
2
2
2
(k) x – 2xy cot – y = 0 (l) x + 2xy tanα + y = 0
2
2
2
2
(m) x + 2xy + y – 2x – 2y – 15 = 0 (n) 2x + 3xy + y + 5x + 2y – 3 = 0

4. Prove that the lines represented by the following equations are perpendicular to each other.
2
2
2
2
(a) 4x – 4y = 0 (b) 5x – 24xy – 5y = 0
2
2
2
2
(c) 4x + 6xy – 4y = 0 (d) 7y – 6xy – 7 x = 0
2
2
2
2
(e) x + 2xy cot – y = 0 (f) y – 2xy tan – x = 0
2
2
2
2
(g) x – y + 2x + 4y – 3 = 0 (h) 12x + 7xy – 12y – 4x + 3y = 0
5. Prove that the lines represented by the following equations are coincident.
2
2
2
2
(a) x + 2xy + y = 0 (b) 4x – 12xy + 9y = 0
2
2
2
2
(c) 5x – 10xy + 5y = 0 (d) 9y + 6xy + x = 0
2
2
2
2
(e) 4x + 4xy + y – 4x – 2y – 3 = 0 (f) 9x + 24xy + 16y – 3x – 4y = 0
2
2
2
2
2
2
(g) x + 2xy cosec + y cosec  = 0 (h) y sin² + 2xy sin. cos + x cos  = 0

6. Find the value of k when the lines represented by the following equations are perpendicular
to each other.
2
2
2
2
(a) kx + 8xy – 3y = 0 (b) 8x + 12xy + ky = 0
2
2
2
2
2
(c) (k – 8) x – 9xy + 3ky = 0 (d) (k – 6)x – 5xy + 5ky = 0
2
2 2
(e) 2k y + 4xy – (3k + 5)x = 0

7. Find the value of m when the lines represented by the following equations are coincident.
2
2
2
2
(a) 4x – 12xy + my = 0 (b) 25x – 20xy + my = 0
2
2
2
2
(c) 9x + mxy + 64y = 0 (d) (m + 2)x + 8xy + 4y = 0
2
2
(e) (m + 1)x – 4mxy + 2y = 0

RAMHARI SHRESTHA [dhadinge@hotmail.com]

2
2
8. Find the two separate equations when the lines represented by kx + 8xy – 3y = 0 are
perpendicular to each other.
2
2
9. Find the two separate equations when the lines represented by 2x – 3xy + ky – x + 2y = 0
are perpendicular to each other.
2
2
10. Find the two separate equations when the lines represented by mx + 12xy + 9y = 0 are
coincident.
2
2
11. Find the two separate equations when the lines represented by x + 2xy + my – 2x – 2y – 3 =
0 are parallel.
12. Find the equations of the straight lines passing through (0, 0) and perpendicular to the pair
2
2
of lines. 2x + 3xy + y + 5x + 2y – 3 = 0
13. Find the equations of the straight lines passing through (1, 2) and perpendicular to the line
2
2
2x + 5xy – 3y = 0.
14. Find the equation of the straight lines passing through (– 1, 4) and parallel to the pair of
2
2
lines 5x – 8xy – 4y = 0.
15. Find the equation of a pair of lines passing through (0, 5) and parallel to the pair of lines
2
2
x – 2xy + y + 2x – 2y = 0.
2
2
16. If  is the acute angle made by the two lines represented by x + 2xy secβ + y = 0, then
show that α = β.
2
2
17. Find the equation of the pair of lines represented by the equation x – 2xy cosec  + y =
0. Also find the angle between them.
2
2
18. Find the equation of the pair of lines represented by the equations 2x + 5xy + 3y = 0 .
Also find the angle between them.

Answers

2
2
1. (a) 15x + 14xy – 8y = 0 3. (a) 45°, 135° 8. x + 3y = 0 and 3x – y = 0
(b) 10x – 31xy + 15y + 5x – 3y (b) 30°, 150°
2
2
= 0 (c) tan (±3) 9. x – 2y = 0 and 2x + y – 1 = 0
–1
(c) 28x – 27 xy – 10y – 21x + (d) °
2
2
37y – 7 = 0 (e) 90° 10. 2x + 3y = 0 and 2x + 3y = 0
(d) mnx – (m – n ) xy – mny (f) 45°, 135°
2
2
2
2
(e) y + 2xy cot – x = 0 (g) 90° 11. x + y + 1 = 0, x + y – 3 = 0
2
2
–1
(f) x + 2xy sec + y = 0 (h) tan (±8)
2
2
2
2
–1
(i) tan (± cot) 12. x – 3xy + 2y = 0
2. (a) x + y = 0 and 3x + 2y = 0 (j) tan (±tan)
–1
2
2
(b) x – 2y = 0 and 4x + 3y = 0 (k) 90° 13. 3x + 5xy – 2y – 16 x + 3y +
(c) 2x + 3y = 0 and 2x – 3y = 0 (l) tan 5 = 0
–1
(d) x = 0 and (x – 5y) = 0, x = 0, x (± tan  – 1)
2
2
2
– 5y = 0 (m) 0° 14. 5x – 8xy – 4y + 42x + 24y –
(e) x = 0 and ax + by = 0  1  27 = 0
–1
(f) x – 3y = 0 and x – 3y – 1 = 0 (n) tan ± 
 3

2
2
(g) x + y = 0 and x – y – 5 = 0 15. x – 2xy + y + 10x – 10y + 25
(h) x + y – 2 = 0 and x + y – 1 = 0 6. (a) 3 = 0
(b) –8
RAMHARI SHRESTHA [dhadinge@hotmail.com]

(i) (x + y + 2) = 0 and x + 2y + 1 (c) 2 17. x– (cosec – cot y = 0
= 0 (d) 1, –6 and x – (cosec + cot)y = 0,
(j) 2x – y + 3 = 0 and x – 3y + 4 5 tan (±cot)
–1
= 0 (e) , –1 18. x + y = 0 and 2x + 3y = 0, tan
2
–1
(k) x – 4 = 0 and y – 5 = 0  1 
(l) y + x(sec + tan) = 0 7. (a) ± 6 ± 

 5
and y + x (sec – tan) = 0 (b) 4
(m) x + y( cosec + cot) = (c) ±48
0 (d) 2
1
and x + y(cosec – cot) = 0 (e) 1, –
(n) x + y (tan + sec) = 0 2
and x + y (tan – sec) = 0
(o) x – y (cot + cosec) = 0
and x – y (cot – cosec) = 0
(p) x – y = 0 and x + y – 1 = 0
(q) ax – by = 0 and bx – ay = 0
(r) x + y + 3 = 0 and x – y – 3 = 0
(s) x + y – 5 = 0 and x + y + 3 = 0
(t) x – y = 0 and 2x – (x + y
)cos  = 0
2
(u) 8x + (3 – –23 )y = 0 & 8x +
(3 + –23 ) = 0

















































RAMHARI SHRESTHA [dhadinge@hotmail.com]