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.

#### Solution

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.

\[\text { Then,} \left| \frac{m - 7}{1 + 7m} \right| = \left| \frac{m + 1}{1 - m} \right| = \left| \frac{m + 1}{m - 1} \right|\]

\[ \Rightarrow \frac{m - 7}{1 + 7m} = \pm \frac{m + 1}{m - 1}\]

Taking the positive sign, we get:

\[m^2 - 8m + 7 = 7 m^2 + 8m + 1\]

\[ \Rightarrow \left( m + 3 \right)\left( m - \frac{1}{3} \right) = 0\]

\[ \Rightarrow m = - 3, \frac{1}{3}\]

Now, taking the negative sign, we get:

\[\left( m - 7 \right)\left( m - 1 \right) = - \left( 7m + 1 \right)\left( m + 1 \right)\]

\[ \Rightarrow m^2 - 8m + 7 = - 7 m^2 - 8m - 1\]

\[ \Rightarrow m^2 = - 1\text { (not possible) }\]

Equations of the third side is

\[y + 10 = - 3\left( x - 1 \right) \text { and } y + 10 = \frac{1}{3}\left( x - 1 \right)\]

\[ \Rightarrow 3x + y + 7 = 0 \text { and } x - 3y - 31 = 0\]