The vertices of a triangle are A(1, 7),B(-5, | vedclass

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Question

(A) The vertices of the triangle are $A(1, 7)$,$B(-5, -1)$,and $C(-1, 2)$.The equation of line $AB$ passing through $(1, 7)$ and $(-5, -1)$ is:$y - 7

Vedclass · Accepted Answer

$x-y+4=0$

Vedclass · Answer

(A) The vertices of the triangle are $A(1, 7)$,$B(-5, -1)$,and $C(-1, 2)$.The equation of line $AB$ passing through $(1, 7)$ and $(-5, -1)$ is:$y - 7 = \frac{-1 - 7}{-5 - 1}(x - 1)$ $\Rightarrow y - 7 = \frac{-8}{-6}(x - 1)$ $\Rightarrow y - 7 = \frac{4}{3}(x - 1)$$3y - 21 = 4x - 4 \Rightarrow 4x - 3y + 17 = 0$.The equation of line $BC$ passing through $(-5, -1)$ and $(-1, 2)$ is:$y - (-1) = \frac{2 - (-1)}{-1 - (-5)}(x - (-5)) \Rightarrow y + 1 = \frac{3}{4}(x + 5)$$4y + 4 = 3x + 15 \Rightarrow 3x - 4y + 11 = 0$.The equation of the angle bisector of $\angle ABC$ is given by:$\frac{4x - 3y + 17}{\sqrt{4^2 + (-3)^2}} = \pm \frac{3x - 4y + 11}{\sqrt{3^2 + (-4)^2}}$$\frac{4x - 3y + 17}{5} = \pm \frac{3x - 4y + 11}{5}$Case $1$: $4x - 3y + 17 = 3x - 4y + 11 \Rightarrow x + y + 6 = 0$.Case $2$: $4x - 3y + 17 = -(3x - 4y + 11)$ $\Rightarrow 4x - 3y + 17 = -3x + 4y - 11$ $\Rightarrow 7x - 7y + 28 = 0$ $\Rightarrow x - y + 4 = 0$.Comparing with the options,$x - y + 4 = 0$ is the correct equation.

The vertices of a triangle are $A(1, 7)$,$B(-5, -1)$,and $C(-1, 2)$. Then,the equation of a bisector of the $\angle ABC$ is

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