From the question
We are given the points
[tex]K(-6,6),P(-3,-2)[/tex]
Finding the slopre, m
Slope is calculated using
[tex]m=\frac{y_{2_{}}-y_1}{x_2-x_1}[/tex]
From the given points
[tex]\begin{gathered} x_1=-6,y_1=6 \\ x_2=-3,y_2=-2 \end{gathered}[/tex]
Therefore,
[tex]\begin{gathered} m=\frac{-2-6}{-3-(-6)} \\ m=\frac{-8}{-3+6} \\ m=\frac{-8}{3} \end{gathered}[/tex]
Therefore, m = -8/3
The next thing is to find
[tex]\mleft\Vert m\mright?[/tex]
A slope parallel to m
For parallel lines, slopes are equal
Therefore,
[tex]\mleft\Vert m=-\frac{8}{3}\mright?[/tex]
Next, we are to find
[tex]\perp m[/tex]
A slope perpendicular to m
For perpendicular lines, the product of the slopes = -1
Therefore
[tex]\perp m=-\frac{1}{m}[/tex]
Hence,
[tex]\begin{gathered} \perp m=-\frac{1}{-\frac{8}{3}} \\ \perp m=\frac{3}{8} \end{gathered}[/tex]
Therefore,
[tex]\perp m=\frac{3}{8}[/tex]
Next, we are to find the distance KP
Using the formula
[tex]KP=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
This gives
[tex]\begin{gathered} KP=\sqrt[]{(-3-(-6))^2+(-2-6)^2} \\ KP=\sqrt[]{3^2+(-8)^2} \\ KP=\sqrt[]{9+64} \\ KP=\sqrt[]{73} \end{gathered}[/tex]
Therefore,
[tex]\text{Distance }=\sqrt[]{73}[/tex]
Next, equation of the line
The equation can be calculated using
[tex]\frac{y-y_1}{x-x_1}=m[/tex]
By inserting values we have
[tex]\begin{gathered} \frac{y-6}{x-(-6)}=-\frac{8}{3} \\ \frac{y-6}{x+6}=-\frac{8}{3} \\ y-6=\frac{-8}{3}(x+6) \\ y-6=-\frac{8}{3}x-6(\frac{8}{3}) \\ y-6=-\frac{8}{3}x-16 \\ y=-\frac{8}{3}x-16+6 \\ y=-\frac{8}{3}x-10 \end{gathered}[/tex]
Therefore the equation is
[tex]y=-\frac{8}{3}x-10[/tex]
