Chapter 3.6, Problem 15PPS

To determine

To find distance from p to l. The line contains points $(0,-3)$ and $(7,4)$ .Point p has coordinates $(4,3)$ .

Answer to Problem 15PPS

  $d=\frac{2}{\sqrt{2}}units$

Explanation of Solution

Given information:

The line contains points $(0,-3)$ and $(7,4)$ .

Point p has coordinates $(4,3)$ .

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 $(0,-3)$ and $(7,4)$ .

Then

  $x_1=0$

  $y_1=-3$

  $x_2=7$

  $y_2=4$

Using slope formula

  $m=\frac{4-(-3)}{7-0}$

On solving

  $m=\frac{7}{7}$

On solving

  $m=1$

Then point slope form

Take any one point that line containing

Let point taken as $(0,-3)$

And $m=1$

So

  $y-y_1=m(x-x_1)$

Substitute values of point and slope

  $y-(-3)=1(x-0)$

On solving

  $y+3=x$

On solving

  $x-y-3=0$

Here

  $a=1$

  $b=-1$

  $c=-3$

And

Point from the line is $(4,3)$

Here $x_p=4$

  $y_p=3$

Using distance formula

Substitute values of line and point

  $d=\frac{|1\times 4+(-1)\times 3-3|}{\sqrt{1^2+{(-1)}^2}}$

On solving

  $d=\frac{|4-3-3|}{\sqrt{1+1}}$

On solving

  $d=\frac{|-2|}{\sqrt{2}}$

On solving

  $d=\frac{2}{\sqrt{2}}units$

Hence, distance from point to line is $\frac{2}{\sqrt{2}}units$ .