Question

If $r(1-m^2)+m(p-q)=0$, then a bisector of the angle between the lines represented by the equation $p{x}^2-2rxy+q{y}^2=0$, is

• A. $y=x$
• B. $y=-x$
• C. $y=mx$
• D. $my=x$

Answer 1

The correct option is C

$y=mx$

Explanation for the correct option.

Step 1: Find angle bisector

Given that, $r(1-m^2)+m(p-q)=0$

$p{x}^2-2rxy+q{y}^2=0$ -----(1)

The general equation of line is $a{x}^2+2hxy+b{y}^2=0$.

Comparing general equation with equation (1) we get:

$a=p,h=-r,b=q$

Angle bisector is given by$\frac{(x^2–y^2)}{\left(a-b\right)}=\frac{xy}{h}$.

$$\begin{gathered} \Rightarrow \frac{(x^2–y^2)}{\left(p-q\right)}=\frac{-xy}{r} \\ \Rightarrow \frac{(x^2–y^2)}{xy}=\frac{-\left(p-q\right)}{r}...\left(2\right) \end{gathered}$$

Step 2: Find equation

Also given $r(1-m^2)+m(p-q)=0$

$-\frac{(p-q)}{r}=\frac{1-m^2}{m}....\left(3\right)$

Substitute equation (3) in (2).

$\frac{x^2-y^2}{xy}=\frac{1-m^2}{m}$

$\begin{array}{rcl}\frac{\left[1-{\left(\frac{y}{x}\right)}^2\right]}{{\displaystyle \frac{y}{x}}}& =& \frac{1-m^2}{m}\\ \left[1-{\left(\frac{y}{x}\right)}^2\right]& =& \left(\frac{1-m^2}{m}\right)\frac{y}{x}\\ \left[\frac{x^2-y^2}{x^2}\right]& =& \left(\frac{y-m^{2}y}{mx}\right)\\ y& =& mx\end{array}$

Hence, option C is correct.