Question

The slope of angular bisectors of pair of lines(ax + by)2 = c (bx - ay), (с > 0) ​

Answer 1

-a/b,b/a is the slope of angular bisector

Answer 2

Answer:

Concept:

When two linear equations that each represent a straight line in (x) and (y) are multiplied together, a pair of straight lines is created.

Step-by-step explanation:

Given:

The bisectors of the angle between the lines

$(a x+b y)^{2}=C(b x-a y)^{2}, c > 0$  are respectively parallel and perpendicular to the line.

Find:

The slope of angular bisectors of pair of given lines

Solution:

Here, $(a x+b y)^{2}=c(b x-a y)^{2}$

$\Rightarrow a^{2} x^{2}+b^{2} y^{2}-2 a b x y=c b^{2} x^{2}+c a^{2} y^{2}+2 a b c x y$

$\Rightarrow x^{2}\left(a^{2}-b^{2} c\right)+y^{2}\left(b^{2}-a^{2} c\right)+(2 a b c+2 a b) x y=0$

on comparing with the standard equation

we get,

$a=\left(a^{2}-b^{2} c\right), b=\left(b^{2}-a^{2} c\right), h=(a b c+a b)$

now equation of the bisector of an angle is given

           $\begin{aligned}&\frac{\left(x^{2}-y^{2}\right)}{x y}=\frac{(a-b)}{b} \\&\frac{\Rightarrow\left(x^{2}-y^{2}\right)}{x y}=\frac{\left[\left(a^{2}-b^{2} c\right)-\left(b^{2}-a^{2}]\right.\right.}{(a b c+a b)} \\&\Rightarrow \frac{\left(x^{2}-y^{2}\right)}{x y}=\frac{\left(a^{2}-b^{2}\right)}{a b} \\&\Rightarrow a b x^{2}-a b y^{2}=\left(a^{2}-b^{2}\right) \times y\end{aligned}$

Let m1 and m2 be the slope of the bisector of an angle.

so, $m_{1}+m_{2}=\frac{-2 h}{b}=\frac{-\left(a^{2}-b^{2}\right)}{a b}=\frac{b}{a}=-\frac{a}{b}$

$m_{1} m_{2}=-1 so, m_{1}=\frac{b}{a} and m_{2}=\frac{-a}{b}$    (b/a , -a/b)

now here slope of the line $a x+b y+k=0$ is $\frac{-a}{b}$

here it is clear that $\frac{-a}{b}=m_{1}$   i.e.,  one of the bisectors of angle is parallel to

line, and $\frac{-a}{b} \times m_{2}=-1$.    so, one of the bisectors is perpendicular to the line.

#SPJ3