Question

Equation of two equal sides of a triangle are the lines $7x+3y-20=0$ and $3x+7y-20=0$ and the third side passes through the point $(-3,3)$ then the equation of the third side can be -

• A. $x+y=0$
• B. $x-y+4=0$
• C. $x+3=0$
• D. $y=3$

Answer 1

The correct option is A $x+y=0$

R.E.F image

Let $AB=AC$

Equation of $AB$ is

$7x+3y-20=0$____(i)

$3x+7y-20=0$_____(ii)

let slope of ${}^{\prime }B{C}^{\prime }$ is ${}^{\prime }{m}^{\prime }$

$\left|\frac{m-m_1}{1+m{m}_1}\right|=\left|\frac{m-m_2}{1+m{m}_2}\right|$

$\Rightarrow \left|\frac{m+\frac{7}{3}}{1-\frac{7m}{3}}\right|=\left|\frac{m+\frac{3}{7}}{1-\frac{3m}{7}}\right|$

$\Rightarrow \left|\frac{3m+7}{3-7m}\right|=\left|\frac{7m+3}{7-3m}\right|$

$\therefore \frac{3m+7}{3-7m}=\pm \frac{(7m+3)}{7-3m}$

Taking $(+)$ sign $m^2=-1$ (Impossible)

Taking $(-)$ sign $m^2=1$

$\therefore m=\pm 1$

Equation of ${}^{\prime }B{C}^{\prime }$ is

$y-3=x+3$

$\Rightarrow y=x+6$

$\Rightarrow x-y+6=0$

when $m=-1$

$y-3=-1(x+3)$

$\Rightarrow x+y=0$