Question

Prove that the equation 2x^2 + 3xy - 2y^2 + 3x + y + 1 represents a pair of perpendicular lines and find the lines.

Answer 1

$\underline{\textsf{To prove:}}$

$\textsf{The equation}\;\mathsf{2x^2+3xy-2y^2+3x+y+1=0}$

$\textsf{represents a pair of perpendicular ines}$

$\underline{\textsf{Solution:}}$

$\textsf{Concept:}$

$\textsf{The condition for the equation}\;\mathsf{ax^2+2hxy+by^2+2gx+2fy+c=0}$

$\textsf{represents a pair of perpendicular line is a+b=0}$

$\textsf{Comparing,}$

$\mathsf{2x^2+3xy-2y^2+3x+y+1=0}\;\textsf{with}$

$\mathsf{ax^2+2hxy+by^2+2gx+2fy+c=0}\;\textsf{we get}$

$\textsf{a=2, h=3/2, b=-2}$

$\textsf{Here, a+b=2+(-2)=0}$

$\therefore\textsf{The given equation represents a pair of perpendicular lines}$

$\textsf{Consider,}$

$\mathsf{2x^2+3xy-2y^2}$

$\mathsf{=2x^2+4xy-xy-2y^2}$

$\mathsf{=2x(x+2y)-y(x+2y)}$

$\mathsf{=(2x-y)(x+2y)}$

$\textsf{Now,}$

$\mathsf{2x^2+3xy-2y^2+3x+y+1=(2x-y+l)(x+2y+m)}$

$\textsf{Equating corresponding coefficients of x and y}$

$\textsf{on bothsides, we get}$

$\mathsf{l+2m=3}$....(1)

$\mathsf{2l-m=1}$......(2)

$\textsf{Solving (1) and (2)}$

$\mathsf{l+2m=3}$

$\mathsf{4l-2m=2}$

$\textsf{Adding,}$

$\mathsf{5\,l=5\implies\;l=1}$

$\mathsf{(2)\implies\;2-m=1\implies\;m=1}$

$\therefore\textsf{The separate equations are}$

$\textsf{2x-y+1=0 and x+2y+1=0}$