Question

Two 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). Determine the equation of the third side.

Answer 1

Let ABC be the isosceles triangle, where 7x − y + 3 = 0 and x + y − 3 = 0 represent the sides AB and AC, respectively.
Let AB = BC

$\because$ AB = BC

$\therefore$ tan B = tan C

Here,

Slope of AB = 7

Slope of AC = −1

Let m be the slope of BC.

$$\begin{gathered} \mathrm{Then,}\left|\frac{m-7}{1+7m}\right|=\left|\frac{m+1}{1-m}\right|=\left|\frac{m+1}{m-1}\right| \\ \Rightarrow \frac{m-7}{1+7m}=\pm \frac{m+1}{m-1} \end{gathered}$$

Taking the positive sign, we get:

$$\begin{gathered} m^2-8m+7=7m^2+8m+1 \\ \Rightarrow \left(m+3\right)\left(m-\frac{1}{3}\right)=0 \\ \Rightarrow m=-3,\frac{1}{3} \end{gathered}$$

Now, taking the negative sign, we get:

$$\begin{gathered} \left(m-7\right)\left(m-1\right)=-\left(7m+1\right)\left(m+1\right) \\ \Rightarrow m^2-8m+7=-7m^2-8m-1 \\ \Rightarrow m^2=-1(\mathrm{not}\mathrm{possible}) \end{gathered}$$

Equations of the third side is

$$\begin{gathered} y+10=-3\left(x-1\right)\mathrm{and}y+10=\frac{1}{3}\left(x-1\right) \\ \Rightarrow 3x+y+7=0\mathrm{and}x-3y-31=0 \end{gathered}$$