Question

If the equation $6x^2-11xy-10y^2-19y+c=0$ represents a pair of lines, find their equations. Also find the angle between the two lines.

Answer 1

Given equations for pair of straight lines is $6x^2-11xy-10y^2-19y+c=0$

$6x^2-11xy-10y^2=19y-c$

$6x^2-15xy+4xy-10y^2=19y-c$

$(3x+2y)(2x-5y)=19y-c$

let the straight lines be $3x+2y+c_1=0,2x-5y+c_2=0$

the pair of lines equations is $6x^2-11xy-10y^2+(2c_1+3c_2)x+(-5c_1+2c_2)y+\left(c_1{c}_2\right)=0$

$⟹2c_1+3c_2=0\cdots \left(1\right)$, $-5c_1+2c_2=-19\cdots \left(2\right)$ and $c=c_1{c}_2$

Solving $\left(1\right)$ and $\left(2\right)$ we get $c_1=3,c_2=-2⟹c=-6$

The pair of straight lines is $6x^2-11xy-10y^2-19y-6=0$

let angle between the pair of straight lines is $\theta$ then $\cos \theta =\frac{\mid \left(3\right)\left(2\right)+\left(2\right)(-5)\mid }{\sqrt{3^2+2^2}\sqrt{2^2+(-5{)}^2}}=\frac{4}{\sqrt{377}}⟹\theta ={\text{cos}}^{-1}(\frac{4}{\sqrt{377}})$