Question

construct a line perpendicular to m through P. Then find the distance from P to m.


Line m contains points (1, 1) and (5,1). point P has coordinates of (2, -4)

Answer 1

Line perpendicular to m is y =–4.

Distance from P to m is 5 units.

Solution:

Line m contains points (1, 1) and (5, 1).

Slope passing through two points formula:

$\text{slope}=\frac{y_2-y_1}{x_2-x_1}$

        $=\frac{1-1}{5-1}$

Slope = 0

Slope of the line perpendicular to the line m:

$\text{Slope}=\frac{-1}{\text{slope}}=0$

Equation of a line passing through one point and slope formula:

$y-y_1=m(x-x_1)$

Here, m = 0 and P(2, –4)

$\Rightarrow y-(-4)=0(x-2)$

$\Rightarrow y+4=0$

$\Rightarrow y=-4$

⇒ y = –4

Equation of a line perpendicular to m and passing through P is y = –4.

Option C is the correct graph. Because it only has slope 0 and P(2, –4).

Point of intersection where line m and P meets is (2, 1).

Let us find the distance between the line m in the point (2, 1) and P(2, –4).

Distance formula:

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

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

              $=\sqrt{25}$

              = 5

Distance = 5 units

Hence line perpendicular to m is y =–4.

Distance from P to m is 5 units.