Question

If $2x^2-5xy+2y^2=0$ represent two equal sides of an isosceles triangle and third side passing through $\left(\begin{array}{cc}3& 3\\ 2& 2\end{array}\right)$

• A. $x+y-1=0$
• B. $x-y+3=0$
• C. $x+y-3=0$
• D. $x+y-6=0$

Answer 1

The correct option is C $x+y-3=0$

Pair of equal sides $\to 2x^2-5xy+2y^2=0$

$\Rightarrow 2x^2-4xy-xy+2y^2=0$

$\Rightarrow 2x(x-2y)-y(x-2y)=0$

$\Rightarrow (x-2y)(2x-y)=0$

$\therefore L_1\Rightarrow x-2y=0$ and $L_2\Rightarrow 2x-y=0$

$m_1=\frac{1}{2}{m}_2=2$

Let slope of third side be $m_3$.

$\because$ it is an isosceles triangle, so third

side will make equal angle with both

of equal sides.

. $\frac{|m_1-m_3|}{|1+m_1{m}_3|}=\frac{|-m_2+m_3|}{|1+m_2{m}_3|}$

$\Rightarrow \frac{|\frac{1}{2}-m_3|}{|1+\frac{m_3}{2}|}=\frac{|-2+m_3|}{|1+2m_3|}$

it is an isoceles triangle so equal angles

need to acute, so tan $\theta >0$

$\Rightarrow \frac{\frac{1}{2}-m_3}{1+\frac{m_3}{2}}=\frac{-2+m_3}{1+2m_3}$

$\Rightarrow \frac{1}{2}+m_3-m_3-2m_3^2=-2+m_3-m_3+\frac{m_3^2}{2}$

$\Rightarrow \frac{1}{2}+2=2m_3^2+\frac{m_3^2}{2}$

$\Rightarrow \frac{5}{2}=\frac{5}{2}{m}_3^2$

$\Rightarrow m_3^2=1\Rightarrow m_3=\pm 1$

So equation of third side $\to \frac{y-\frac{3}{2}}{x-\frac{3}{2}}=1$

$\Rightarrow y-\frac{3}{2}=x-\frac{3}{2}\Rightarrow y=x$

$ar$

$\frac{y-\frac{3}{2}}{x-\frac{3}{2}}=-1\Rightarrow y-\frac{3}{2}=\frac{3}{2}-x$

$\Rightarrow x+y-3=0$

$\therefore$ Answer: option (c)