Question

Two equal sides of an isosceles triangle are given by the equations $7x-y+3=0$ and $x+y-3=0$ and its third side passes through the point $(1,-10)$. The equation of the third side can be

• A. $x+3y+29=0$
• B. $x-3y=31$
• C. $3x-y=13$
• D. $3x+y+7=0$

Answer 1

Answer: (B), (D)

$m_{AB}=7$ and $m_{AC}=-1$
Let slope of $BC$ (third side) be $m$.
Now $\left|\frac{7-m}{1+7m}\right|=\left|\frac{-1-m}{1-m}\right|\Rightarrow \left(\frac{7-m}{1+7m}\right)=\pm \left(\frac{-1-m}{1-m}\right)$

$\therefore$ Taking ' + ' sign, we get
$m^2+1=0$ (no real values of $m$ possible)
And taking'-' sign, we get
$3m^2+8m-3=0\Rightarrow (3m-1)(m+3)=0\Rightarrow m=\frac{1}{3},-3$
$\therefore$ Required equation of third side can be $x-3y-31=0$ or $3x+y+7=0$ Hence, $a=3,b=1$
$\therefore \left(a^2-b^2\right)=9-1=8$.