Prove that the equation $6 x^{2} + 17 x y + 12 y^{2} + 22 x + 31 y + 20 = 0$ represents a pair of straight lines and find their equations.

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Solution

Method to find the two factors of the given equation.
Rule: First factorize the second degree terms of the given equation and then add p and q to each of these factors respectively. Now multiply these two factors, compare the coefficients of x, y and constant in this product with the corresponding coefficients in the given equation. Thus you will find the values of p and q.
$6 x^{2} + 17 x y + 12 y^{2} = 6 x^{2} + 8 x y + 9 x y + 12 y^{2}$
$= 2 x (3 x + 4 y) +) 3 y (3 x + 4 y)$
$= (2 x + 3 y) (3 x + 4 y)$ .
Hence the factors of the given equation are $(2 x + 3 y + p) (3 x + 4 y + q)$ . $(1)$
Comparing the coefficients of x, y and constant in the product of $(1)$ with the corresponding coefficients in the given equation, we get
$3 p + 2 q = 22$ .. $(2)$
$4 p + 3 q = 31$ .. $(3)$
and $p q = 20$ $(4)$
On solving $(2)$ and $(3)$ , we find that $p = 4$ and $q = 5$ and these values of p and q satisfy $(4)$ as well. Hence the given equation represents a pair of straight lines whose separate equations are $2 x + 3 y + 4 = 0$ and $43 x + 4 y + 5 = 0$ .

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