Chapter 3.6, Problem 18PPS

To determine

To find distance from p to l. The line contains points $(4,-1)$ and $(4,9)$ .Point p has coordinates $(1,6)$ .

Answer to Problem 18PPS

  $d=3units$

Explanation of Solution

Given information:

The line contains points $(4,-1)$ and $(4,9)$ .

Point p has coordinates $(1,6)$ .

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 $(4,-1)$ and $(4,9)$ .

Then

  $x_1=4$

  $y_1=-1$

  $x_2=4$

  $y_2=9$

Using slope formula

  $m=\frac{9-(-1)}{4-4}$

On solving

  $m=\frac{10}{0}$

Then point slope form

Take any one point that line containing

Let point taken as $(4,-1)$

And $m=\frac{10}{0}$

So

  $y-y_1=m(x-x_1)$

Substitute values of point and slope

  $y-(-1)=\frac{10}{0}(x-4)$

On solving

  $0=10(x-4)$

On solving

  $0=(x-4)$

On solving

  $x-4=0$

Here

  $a=1$

  $b=0$

  $c=-4$

And

Point from the line is $(1,6)$

Here $x_p=1$

  $y_p=6$

Using distance formula

Substitute values of line and point

  $d=\frac{|1\times 1+0\times 6-4|}{\sqrt{1^2+{(0)}^2}}$

On solving

  $d=\frac{|1+0-4|}{\sqrt{1}}$

On solving

  $d=\frac{|-3|}{1}$

On solving

  $d=3units$

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