Chapter 3.6, Problem 16PPS

To determine

To find distance from p to l. The line contains points $(11,-1)$ and $(-3,-11)$ .Point p has coordinates $(-1,1)$ .

Answer to Problem 16PPS

  $d=\frac{46}{\sqrt{74}}units$

Explanation of Solution

Given information:

The line contains points $(11,-1)$ and $(-3,-11)$ .

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

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 $(11,-1)$ and $(-3,-11)$ .

Then

  $x_1=11$

  $y_1=-1$

  $x_2=-3$

  $y_2=-11$

Using slope formula

  $m=\frac{-11-(-1)}{-3-11}$

On solving

  $m=\frac{10}{-14}$

On solving

  $m=\frac{5}{-7}$

Then point slope form

Take any one point that line containing

Let point taken as $(11,-1)$

And $m=\frac{5}{-7}$

So

  $y-y_1=m(x-x_1)$

Substitute values of point and slope

  $y-(-1)=\frac{5}{-7}(x-11)$

On solving

  $y+1=\frac{5}{-7}x+\frac{55}{7}$

On solving

  $y+1-1=\frac{5}{-7}x+\frac{55}{7}-1$

On solving

  $y=\frac{5}{-7}x+\frac{48}{7}$

On solving

  $7y=-5x+48$

On solving

  $-5x-7y+48=0$

Here

  $a=-5$

  $b=-7$

  $c=48$

And

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

Here $x_p=-1$

  $y_p=1$

Using distance formula

Substitute values of line and point

  $d=\frac{|-5\times -1+(-7)\times 1+48|}{\sqrt{{(-5)}^2+{(-7)}^2}}$

On solving

  $d=\frac{|5-7+48|}{\sqrt{25+49}}$

On solving

  $d=\frac{\left|46\right|}{\sqrt{74}}$

On solving

  $d=\frac{46}{\sqrt{74}}units$

Hence, distance from point to line is $\frac{46}{\sqrt{74}}units$