Question

lf the equation $2x^2-5xy+2y^2=0$ represents two sides of an isosceles triangle then the equation of the third side passing through the point $(3,3)$ is

• A. $x+y=3$
• B. $x-y=0$
• C. $2x-y=3$
• D. $x+y-6=0$

Answer 1

Answer: (D)

$x+y-6=0$

$2x^2-5x+2y^2=0$

Substitute $t=\frac{x}{y}$

$2t^2-5t+2=0$

$\Rightarrow t=\frac{5\pm \sqrt{25-16}}{4}$

$\Rightarrow t=\frac{5\pm 3}{4}$

$\Rightarrow t=2,\frac{1}{2}$

$\Rightarrow y=\frac{x}{2},y=2x$

As given

$\tan \theta =\left|\frac{2-\frac{1}{2}}{1+1}\right|=\frac{3}{4}$

Angle bisector of $y=\frac{x}{2}$ and $y=2x$ are

$\left|\frac{2y-x}{\sqrt{5}}\right|=\left|\frac{y-2x}{\sqrt{5}}\right|$

$\Rightarrow 2y-x=y-2x$ or
$\Rightarrow 2y-x=2x-y$

$\Rightarrow y=3x$ and $y=x$

The point $(3,3)$ lies one of the angle bisectors of verticle angle of isosceles triangle i.e. $y=x$

The equation of $3^{rd}$ side having slope $(-1)$ is,

$y-3=-(x-3)$

$\Rightarrow y+x=6$