Chapter 3.6, Problem 17PPS

To determine

To find distance from p to l. The line contains points $(-2,1)$ and $(4,1)$ .Point p has coordinates $(5,7)$ .

Answer to Problem 17PPS

  $d=6units$

Explanation of Solution

Given information:

The line contains points $(-2,1)$ and $(4,1)$ .

Point p has coordinates $(5,7)$ .

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 $(-2,1)$ and $(4,1)$ .

Then

  $x_1=-2$

  $y_1=1$

  $x_2=4$

  $y_2=1$

Using slope formula

  $m=\frac{1-1}{4-(-2)}$

On solving

  $m=0$

Then point slope form

Let take any one point that line containing

Let point taken as $(-2,1)$

  $(-2,1)$

And $m=0$

So

  $y-y_1=m(x-x_1)$

Substitute values of point and slope

  $y-1=0(x-(-2))$

On solving

  $y-1=0$

Here

  $a=0$

  $b=1$

  $c=-1$

And

Point $(5,7)$

Here $x_p=5$

  $y_p=7$

Using distance formula

Substitute values of line and point

  $d=\frac{|0\times 5+1\times 7-1|}{\sqrt{0^2+1^2}}$

On solving

  $d=\frac{|0+7-1|}{\sqrt{1}}$

On solving

  $d=\frac{\left|6\right|}{1}$

On solving

  $d=6units$

Hence, distance from point to line is $6units$