If the given equation is in the form
ax^{2} + 2hxy + by^{2} = 0
by factoring we may get two linear factors and that will be the separate equations for the straight line.
If the given equation is in the form
ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0
then we have to take the first three terms and find factors.
Let us see an example problem for better understanding.
Question 1 :
Find the separate equation of the following pair of straight lines
(i) 3x^{2} + 2xy − y^{2} = 0
Solution :
3x^{2} - xy + 3xy − y^{2} = 0
x(3x - y) + y(3x - y) = 0
(3x - y) (x + y) = 0
(ii) 6(x − 1)^{2} + 5(x − 1)(y − 2) − 4(y − 2)^{2} = 0
Solution :
6(x − 1)^{2} + 5(x − 1)(y − 2) − 4(y − 2)^{2} = 0
6(x − 1)^{2} - 3(x − 1)(y − 2) + 8(x − 1)(y − 2) − 4(y − 2)^{2} = 0
3(x- 1) [2(x- 1) - (y -2)] + 4(y - 2) [2(x - 1) - (y - 2)] = 0
3(x- 1) [2x - 2 - y + 2] + 4(y - 2) [2x - 2 - y + 2] = 0
3(x- 1) (2x- y) + 4(y - 2) (2x - y) = 0
(2x - y) (3x - 3 + 4y - 8) = 0
(2x - y) (3x + 4y - 11) = 0
(iii) 2x^{2} − xy − 3y^{2} − 6x + 19y − 20 = 0
Solution :
= 2x^{2} − xy − 3y^{2}
= 2x^{2} + 2xy - 3xy − 3y^{2}
= 2x (x + y) - 3y(x + y)
= (2x - 3y) (x + y)
Separate equations of the straight line are
(2x - 3y + l) (x + y + m)
In order to find the values of l and m, we have to equate the coefficients of x and y.
2x^{2} − xy − 3y^{2} − 6x + 19y − 20 = (2x - 3y + l) (x + y + m)
-6 = 2m + l ------------(1)
19 = -3m + l ------------(2)
(1) - (2)
(2m + l) - (-3m + l) = -6 - 19
2m + l + 3m - l = -25
5m = -25 ==> m = -5
By applying the value of m in the first equation, we get
2(-5) + l = -6
-10 + l = -6
l = -6 + 10 ==> 4
Hence the required equations are (2x - 3y + 4) (x + y - 5).
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