Question

Let $u\equiv ax+by+a\sqrt[3]{3b}=0,v\equiv bx-ay+b\sqrt[3]{3a}=0,a,bϵR$ be two straight lines. The equation of the bisection of the angle formed by $k_{1}u-k_{2}v=0$ & $k_{1}u+k_{2}v=0$ for non zero real $k_1$ & $k_2$ is

• A. $k=1$
• B. $k=3$
• C. $k=2$
• D. $k=4$

Answer 1

The correct option is A $k=1$

Given Lines

$u\equiv ax+by+a\sqrt[3]{3b}=0$

$v\equiv bx-ay+b\sqrt[3]{3a}=0$

Equation of angle bisector

$ax+by+a\sqrt[3]{3b}=\pm bx-ay+b\sqrt[3]{3a}$

$x(a-b)+y(b+a)+a\sqrt[3]{3b}-b\sqrt[3]{3a}=0$ and $x(a+b)+y(b-a)+a\sqrt[3]{3b}+b\sqrt[3]{3a}=0$

The eq of bisector also formed by $k_{1}u-k_{2}v=0$ and $k_{1}u+k_{2}v=0$

Hence

$k_{1}u-k_{2}v=x(a-b)+y(b+a)+a\sqrt[3]{3b}-b\sqrt[3]{3a}$

$k_1(ax+by+a\sqrt[3]{3b})-k_2(bx-ay+b\sqrt[3]{3a})=x(a-b)+y(b+a)+a\sqrt[3]{3b}-b\sqrt[3]{3a}$

$x(k_{1}a-k_{2}b)+y(k_{1}b+k_{2}a)+k_{1}a\sqrt[3]{3b}-k_{2}b\sqrt[3]{3a}=x(a-b)+y(b+a)+a\sqrt[3]{3b}-b\sqrt[3]{3a}$

On comaparing both sides

$k_1=k_2=1$