Question

For points K (-6,6) and P (-3,-2), find the following:m:| m:I m:Distance:Equation of a line 1:

Answer 1

From the question

We are given the points

$K(-6,6),P(-3,-2)$

Finding the slopre, m

Slope is calculated using

$m=\frac{y_{2_{}}-y_1}{x_2-x_1}$

From the given points

$\begin{gathered} x_1=-6,y_1=6 \\ x_2=-3,y_2=-2 \end{gathered}$

Therefore,

$\begin{gathered} m=\frac{-2-6}{-3-(-6)} \\ m=\frac{-8}{-3+6} \\ m=\frac{-8}{3} \end{gathered}$

Therefore, m = -8/3

The next thing is to find

$\mleft\Vert m\mright?$

A slope parallel to m

For parallel lines, slopes are equal

Therefore,

$\mleft\Vert m=-\frac{8}{3}\mright?$

Next, we are to find

$\perp m$

A slope perpendicular to m

For perpendicular lines, the product of the slopes = -1

Therefore

$\perp m=-\frac{1}{m}$

Hence,

$\begin{gathered} \perp m=-\frac{1}{-\frac{8}{3}} \\ \perp m=\frac{3}{8} \end{gathered}$

Therefore,

$\perp m=\frac{3}{8}$

Next, we are to find the distance KP

Using the formula

$KP=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}$

This gives

$\begin{gathered} KP=\sqrt[]{(-3-(-6))^2+(-2-6)^2} \\ KP=\sqrt[]{3^2+(-8)^2} \\ KP=\sqrt[]{9+64} \\ KP=\sqrt[]{73} \end{gathered}$

Therefore,

$\text{Distance }=\sqrt[]{73}$

Next, equation of the line

The equation can be calculated using

$\frac{y-y_1}{x-x_1}=m$

By inserting values we have

$\begin{gathered} \frac{y-6}{x-(-6)}=-\frac{8}{3} \\ \frac{y-6}{x+6}=-\frac{8}{3} \\ y-6=\frac{-8}{3}(x+6) \\ y-6=-\frac{8}{3}x-6(\frac{8}{3}) \\ y-6=-\frac{8}{3}x-16 \\ y=-\frac{8}{3}x-16+6 \\ y=-\frac{8}{3}x-10 \end{gathered}$

Therefore the equation is

$y=-\frac{8}{3}x-10$