Chapter 3.6, Problem 20PPS

To determine

To find distance from p to l. The line contains points $(-8,1)$ and $(3,1)$ .Point p has coordinates $(-2,4)$ .

Answer to Problem 20PPS

  $d=3units$

Explanation of Solution

Given information:

The line contains points $(-8,1)$ and $(3,1)$ .

Point p has coordinates $(-2,4)$ .

Formula used:

Distance from point to line

  $d=\frac{|a{x}_p+b{y}_p+c|}{\sqrt{a^2+b^2}}$

Where

  $a$ , $b$ and $c$ are coefficient of line.

  $x_p$ and $y_p$ are x and y coordinates of point.

Slope of line, $m=\frac{y_2-y_1}{x_2-x_1}$

Calculation:

The line contains points $(-8,1)$ and $(3,1)$ .

Then

  $x_1=-8$

  $y_1=1$

  $x_2=3$

  $y_2=1$

Using slope formula

  $m=\frac{1-1}{3-(-8)}$

On solving

  $m=\frac{0}{11}$

On solving

  $m=0$

Then point slope form

Take any one point that line containing

Let point taken as $(-8,1)$

And $m=0$

So

  $y-y_1=m(x-x_1)$

Substitute values of point and slope

  $y-1=0(x-(-8))$

On solving

  $y-1=0$

Here

  $a=0$

  $b=1$

  $c=-1$

And

Point from the line is $(-2,4)$

Here $x_p=-2$

  $y_p=4$

Using distance formula

Substitute values of line and point

  $d=\frac{|0\times -2+1\times 4-1|}{\sqrt{0^2+{(1)}^2}}$

On solving

  $d=\frac{|0+4-1|}{\sqrt{1}}$

On solving

  $d=\frac{\left|3\right|}{1}$

On solving

  $d=3units$

Hence, distance from point to line is $3units$