Question

Find the joint equation of pair of lines passing through (1,1) and making an angle of ${45}^o$ with the line $3x-4y+1=0$

Answer 1

Let a line passing through $(1,1)$ be

$y=mx+c$

$1=m+c$

$c=1-m$

$y=mx+1-m$

Another line be $y=nx+c$ through $(1,1)$

$c=1-n$

$y=nx+1-n$

Now $mn=-1$ (perpendicular lines)

angle bisector is $3x-4y+1=0$

$y=\frac{3x+1}{4}$

Now $\frac{m+n}{2}=\frac{3}{4}$

Put $m=-\frac{1}{n}$

$\frac{-\frac{1}{n}+n}{2}=\frac{3}{4}$

$-1+n^2=\frac{3}{2}n$

$2n^2-3n-2=0$

$2n^2-4n+n-2=0$

$2n(n-2)+1(n-2)=0$

$n=-\frac{1}{2},2$ $m=2,-\frac{1}{2}$

Lines are $y=-\frac{1}{2}x+\frac{3}{2}$

$2y+x-3=0$ and $y=2x-1$ $y-2x+1=0$

Joint equation

$-2x^2-3xy+2y^2+7x-y+3=0$