Question

Let $u\equiv ax+by+a\sqrt[3]{3b}=0$, $v\equiv bx-ay+b\sqrt[3]{3a}=0$ where a, $b\in R-0$ 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 = 2
• B. $k_{2}u+k_{1}v=0$
• C. $k_{2}u-k_{1}v=0$
• D. v = 0, u=0

Answer 1

The correct option is D v = 0, u=0

Note that the lines are perpendicular.

Assume the co-ordinate are to be directed along $u=0$ & $v=0$.

Now the lines $k_{1}u-k_{2}v=0$ & $k_{1}u+k_{2}v=0$ are equally inclined with $u-v$ axes.

Hence the bisectors are $u=0$ & $v=0$.