Question

Let $u\equiv ax+by+a\sqrt[3]{3b}=0,v=bx-ay+b\sqrt[3]{3a}=0$, $a,b,$ $\in$ R be two straight lines. The equations of the bisectors of the angle formed by $k_{1}u-k_{2}v=0$ and $k_{1}u+k_{2}v=0$ for nonzero real $k_1$ and $k_2$ 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), (D)

The correct options are
A $u=0$
D $v=0$

Since the lines u=0 and v=0 are perpendicular, we can assume that the coordinate axes to be directed towards u=0 and v=0. Since the lines $k_{1}u+k_{2}v=0$ and $k_{1}u+k_{2}v=0$ are equally inclined with the new uv axises similar to the lines $y=x+c$ and $y=-x+c$ which are equally inclined with the xy axises. Hence from the above deductions, we can say that the angle bisectors of the lines
$k_{1}u+k_{2}v=0$ and $k_{1}u+k_{2}v=0$ are $u=0$ and $v=0$