Chapter 3.6, Problem 5CYU

To determine

To Find: The distance from $P$ to $l$ .

Answer to Problem 5CYU

  $\sqrt{10}$

Explanation of Solution

Given:

Line $l$ contains points (-6, 1) and (9, -4). Point $P$ has coordinates (4, 1).

Formula Used:

1. Slope of a line passing through two points ( $x_1,y_1$ ) and ( $x_2,y_2$)

$m=\frac{y_2-y_1}{x_2-x_1}$

2. Slope intercept form of a straight line:

$y=mx+c$

3. If slopes of two perpendicular lines are $m_1,m_2$, then

$m_1.m_2=-1$

4. Distance formula:

$D=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

( $x_1,y_1$ ) = (-6, 1) and ( $x_2,y_2$ ) = (9, -4) Slope of the line $l$ :

  $m=\frac{-4-1}{9-(-6)}=-\frac{1}{3}$

The equation of line $l$ becomes:

  $y=-\frac{1}{3}x+c$

Point (-6, 1) will lie on the line.

Putting (-6, 1) to find the value of $c$ :

  $1=-\frac{1}{3}(-6)+c\Rightarrow c=-1$

Now, equation of line $l$ can be written as:

  $y=-\frac{1}{3}x-1$ ...... (1)

To find the distance of point $P$ to the line $l$ , the equation of line perpendicular to line $l$ has to be found.

Product of slopes of two perpendicular lines is equal to -1.

Let the slope of perpendicular line be $m_1$ .

  $m_1\times -\frac{1}{3}=-1\Rightarrow m_1=3$

Equation of line perpendicular to line $l$ :

  $y=3x+c_1$

Point $P$ (4, 1) will lie on this line.

Putting (4, 1) to find the value of $c_1$ :

  $1=3(4)+c_1\Rightarrow c_1=-11$

Equation of line perpendicular to line $l$ :

  $y=3x-11$ .....(2)

Solving the equation of two lines to find the intersection point of line $l$ and the line perpendicular to line $l$ .

  $x=3,y=-2$

Now, using the distance formula to find the distance between point $P$ and line $l$ .

  $D=\sqrt{{(4-3)}^2+{(1-(-2))}^2}=\sqrt{10}$