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=(-2-6)/(-3-(-6)) \\ m=(-8)/(-3+6) \\ m=(-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=-(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=-(1)/(m)`

Hence,

`\begin{gathered} \perp m=-(1)/(-(8)/(3)) \\ \perp m=(3)/(8) \end{gathered}`

Therefore,

`\perp m=(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

`(y-y_1)/(x-x_1)=m`

By inserting values we have

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

Therefore the equation is

`y=-(8)/(3)x-10`