Chapter 3.6, Problem 4CYU

To determine

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

Answer to Problem 4CYU

  $3\sqrt{5}$

Explanation of Solution

Given:

Line $l$ contains points (4, 3) and (-2, 0). Point $P$ has coordinates (3, 10).

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:

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

( $x_1,y_1$ ) = (3, 4) and ( $x_2,y_2$ ) = (-2, 0) Slope of the line $l$ :

  $m=\frac{0-3}{-2-4}=\frac{1}{2}$

The equation of line $l$ becomes:

  $y=\frac{1}{2}x+c$

Point (4, 3) will lie on the line.

Putting (4, 3) to find the value of $c$ :

  $3=\frac{1}{2}(4)+c\Rightarrow c=1$

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

  $y=\frac{1}{2}x+1$ ...... (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 $m_1$ .

  $m_1\times \frac{1}{2}=-1\Rightarrow m_1=-2$

Equation of line perpendicular to line $l$ :

  $y=-2x+c_1$

Point $P$ (3, 10) will lie on this line.

Putting (3, 10) to find the value of $c_1$ :

  $10=-2(3)+c_1\Rightarrow c_1=16$

Equation of line perpendicular to line $l$ :

  $y=-2x+16$ ..... (2)

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

  $x=6,y=4$

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

  $D=\sqrt{{(6-3)}^2+{(4-10)}^2}=3\sqrt{5}$