Given vertices $A (1, 1), B (4, - 2)$ and $C (5, 5)$ of a triangle, find the equation of the perpendicular dropped from $C$ to the interior bisector of the angle $A$ .

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Solution

The internal bisector of the angle $A$ will divide the opposite side $B C$ at $D$ in the ratio of arms of the angle ie $A B = 3 \sqrt{2}; A C = 4 \sqrt{2}$ Hence by ratio formula the point $D$ is $(\frac{31}{7}, 1)$
Slope of $A D$ by $\frac{y_{2} - y_{1}}{x_{2} - x_{1}} = 0$
Slope of a line perpendicular to $A D$ is $\infty$ . Any line through $C$ perpendicular to this bisector is
$\frac{y - 5}{x - 5} = m = \infty$
$∴$ $x - 5 = 0$

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