Question

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

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

Answer 1

The correct option is A $3x+y+7=0,x-3y-31=0$

$L_1:7x-y+3=0$$m_1=7$

$L_2:x+y-3=0$$m_2=-1$

In an isosceles triangle, the unequal side makes equal angle with sides which are equal. Also equal angles need to be acute.

Let slope of third side be $m$

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

$\Rightarrow \frac{m-7}{1+7m}=\frac{-1-m}{1-m}$

$\frac{m-7}{1+7m}=\frac{m+1}{m-1}$

$\Rightarrow m^2-8m+7=7m^2+8m+1$

$\Rightarrow 3m^2+8m-3=0$

$\Rightarrow 3m^2+9m-m-3=0$

$\Rightarrow 3m(m+3)-1(m+3)=0$

$m=-3$ and $m=\frac{1}{3}$

we will get two different lines which will get third side.

$y+10=-3\Rightarrow y+10=-3x+3$

$x-1\Rightarrow 3x+y+7=0$

$\frac{y+10}{x-1}=\frac{1}{3}\Rightarrow 3y+30=x-1$

Required lines are $3x+y+7=0$

$3y-x+31=0$

Answer: option (A)