Question

The combined equation of the bisectors of the angle between the lines represented by $\sqrt{3}(x^2+y^2)=4xy$.

• A. $y^2-x^2=0$
• B. $xy=0$
• C. $x^2+y^2=2xy$
• D. None of these

Answer 1

The correct option is A

$y^2-x^2=0$

Explanation of correct answer :

Finding combined equation of bisectors of angle :

Given, $\sqrt{3}(x^2+y^2)=4xy$

Simplifying the equation,

$$\begin{gathered} \Rightarrow \sqrt{3}{x}^2+\sqrt{3}{y}^2=4xy \\ \Rightarrow \sqrt{3}{x}^2+\sqrt{3}{y}^2-4xy=0 \\ \Rightarrow \sqrt{3}{x}^2-3xy-xy+\sqrt{3}{y}^2=0 \\ \Rightarrow \sqrt{3}x(x-\sqrt{3}y)-y(x-\sqrt{3}y)=0 \\ \Rightarrow \left(\sqrt{3}x-y\right)\left(x-\sqrt{3}y\right)=0 \end{gathered}$$

Thus, $y=\sqrt{3}x\text{or}y=\frac{1}{\sqrt{3}}x$

therefore, $y=x\tan 60°\text{or}y=x\tan 30°\left[\tan 60°=\sqrt{3}\mathrm{and}\tan 30°=\frac{1}{\sqrt{3}}\right]$

Bisector angles of X-axis $=\frac{90°}{2}and\left(90°+\frac{90°}{2}\right)$

$=45°\text{and}135°$

Bisectors will be making $45°$ and $135°$ with the x-axis as it divides the $1\text{st}$ quadrant and $2\text{nd}$ quadrant in two equal halves.

So the equation of the bisectors are $y=x\tan 45°$and $y=x\tan 135°$

Combining the equations,

$$\begin{gathered} \Rightarrow \left(y-x\tan 45°\right)(y-x\tan 135°)=0 \\ \Rightarrow (y-x)(y+x)=0 \\ \Rightarrow (y^2-x^2)=0 \end{gathered}$$

Hence, the required equation is $(y^2-x^2)=0$.

Hence , the correct answer is Option (A).