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 $(1,-10)$. Equation of the third side can be

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

Answer 1

Answer: (A), (B)

The correct options are
A $3x+y+7=0$
B $x-3y-31=0$

Given sides of isosceles triangle

$7x-y+3=0$--------(1)

$x+y-3=0$--------(2)

On comparing Both equation with $y=mx+c$ where m is slope

$m_1=7$ and $m_2=-1$

Let the slope of third sides is $m_3=m$

Angle between the two sides can be calculated by $\frac{m_1-m_2}{1+m_1{m}_2}$

Angle between (1) and (3) is equal to the angle between (2) and (3) because it is isosceles triangle

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

On solving above equation we get

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

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

Eq of third line passing through Point $(1,-10)$ and slope $m=-3,\frac{1}{3}$

$y+10=-3(x-1)$ and $y+10=\frac{1}{3}(x-1)$

$y+10=-3x+3$ and $3(y+10)=x-1$

$3x+y+7=0$ and $3y+30=x-1$

$3x+y+7=0$ and $x-3y-31=0$