Question

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

• A. $u=0$
• B. $k_{2}u+k_{1}v=0$
• C. $k_{2}u-k_{1}v=0$
• D. $v=0$

Answer 1

Answer: (A)

$u=0$

Given lines

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

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

Equation of angle bisector be

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

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

$ax+by+a\sqrt[3]{3b}-(bx-ay+b\sqrt[3]{3a})=0$ and $ax+by+a\sqrt[3]{3b}+(bx-ay+b\sqrt[3]{3a})=0$

$u-v=0$------(1) and $u+v=0$------(2)

On comparing both eq with given eq $k_{1}u-k_{2}v=0$ and $k_{1}u+k_{2}v=0$

Hence $k_1=1$ and $k_2=1$

On solving eq (1) and (2)

we get $u=0$