Question

the vertices of a triangle abc are (1,1), (4, - 2) and (5, 5), respectively. the equation of the perpendicular dropped from c to the internal bisector of ∠a is (a) y = 5 (b) x = 5 (c) 2x+3y = 7 (d) none

Explanation:

Step1: Find the angle - bisector formula

The formula for the angle - bisector of $\angle A$ of a triangle with vertices $A(x_1,y_1)$, $B(x_2,y_2)$ and $C(x_3,y_3)$ is given by $\frac{AB}{AC}\cdot\frac{(x - x_1)}{|AB|}+\frac{AC}{AB}\cdot\frac{(x - x_1)}{|AC|}=0$. First, find the lengths of $AB$ and $AC$.
The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.
$AB=\sqrt{(4 - 1)^2+(-2 - 1)^2}=\sqrt{9 + 9}=\sqrt{18}=3\sqrt{2}$
$AC=\sqrt{(5 - 1)^2+(5 - 1)^2}=\sqrt{16 + 16}=\sqrt{32}=4\sqrt{2}$

The vectors $\overrightarrow{AB}=(4 - 1,-2 - 1)=(3,-3)$ and $\overrightarrow{AC}=(5 - 1,5 - 1)=(4,4)$.
The unit - vector along $\overrightarrow{AB}$ is $\hat{u}_{AB}=\frac{\overrightarrow{AB}}{|AB|}=(\frac{3}{3\sqrt{2}},\frac{-3}{3\sqrt{2}})=(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}})$
The unit - vector along $\overrightarrow{AC}$ is $\hat{u}_{AC}=\frac{\overrightarrow{AC}}{|AC|}=(\frac{4}{4\sqrt{2}},\frac{4}{4\sqrt{2}})=(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$
The vector along the angle - bisector of $\angle A$ is $\hat{u}_{AB}+\hat{u}_{AC}=(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}})=(\sqrt{2},0)$
The equation of the line passing through $A(1,1)$ with direction vector $(\sqrt{2},0)$ is $y - 1 = 0(x - 1)$ (since the slope $m = 0$), i.e., $y = 1$.

Step2: Find the perpendicular line

The line perpendicular to $y = 1$ passing through $C(5,5)$:
The slope of the line $y = 1$ is $m_1 = 0$. The slope of a line perpendicular to a line with slope $m_1$ is undefined.
The equation of a line with undefined slope passing through the point $(x_0,y_0)=(5,5)$ is of the form $x=x_0$.
So the equation of the line is $x = 5$.

Answer:

(B) $x = 5$