Question

lf the equation $6x^2+5xy+b{y}^2+9x+20y+c=0$ represents a pair of perpendicular lines, then $b-c=$

• A. $-6$
• B. $-3$
• C. $-2$
• D. $0$

Answer 1

Answer: (D)

$0$

$6x^2+5xy+b{y}^2+9x+20y+c=0$ ----(1)

$\Rightarrow a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$ ----(2)

By comparing 1 & 2

$a=6,b=b$

As given pair of perpendicular lines

i.e. $a+b=0$

$a=-b$

$b=-a=-6$

$b=-6$

Let,

$6x^2+5xy-6y^2=0$

$t=\frac{x}{y},$

$6t^2+5t-6=0$

$t=\frac{-3}{2},\frac{2}{3}$

$2x+3y=0$ and
$3x-2t=0$

Let $(2x+3y+c_1)=0$ and $3x-2y+c_2=0$

Then $(2x+3y+c_1)(3x-2y+c_2)=0$

$(6x^2+5xy-6y^2)+(2c_2+3c_1)x+(3c_2-2c_1)y+c_1{c}_2=0$ ----(3)

$\therefore$ By comparing 1 & 3,

$2c_2+3c_1=9$ ----(4)

$3c_2-2c_1=20$ ---(5)

From 4 & 5,

$c_2=\frac{78}{13}=6$ & $c_1=-1$

$\therefore c=c_1{c}_2=-6$

$\therefore b-c=-6+6=0$