Question

One of the bisector of the angle between the lines $a(x-1{)}^2+2h(x-1\left)\right(y-2)+b(y-2{)}^2=0$ is $x+2y-5=0$ and $\frac{b-a}{h}=\frac{p}{q}$, where $p$ & $q$ are co-prime numbers, then $(p+q)$ is equal to

Answer 1

The given equation is $a(x-1{)}^2+2h(x-1\left)\right(y-2)+b(y-2{)}^2=0$

Let $X=(x-1)$ and $Y=(y-2)$

Equation of angle bisectors:

$\frac{X^2-Y^2}{a-b}=\frac{XY}{h}$

and $x+2y-5=0\Rightarrow X+2Y=0\Rightarrow X=-2Y$

Put this in the above equation to get,

$\frac{b-a}{h}=\frac{3}{2}\Rightarrow p=3,q=2$

$p+q=5$