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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.
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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
\[y + 10 = - 3\left( x - 1 \right) \text { and } y + 10 = \frac{1}{3}\left( x - 1 \right)\]
\[ \Rightarrow 3x + y + 7 = 0 \text { and } x - 3y - 31 = 0\]
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