Perpendicular Lines and its Application

by Dan · Published November 27, 2013 · Updated December 26, 2014

Perpendicular lines are two intersecting lines that meet at right angle or exactly 90°. Let’s bring that line now to coordinate plane. In Analytical Geometry, two lines are perpendicular if the product of their slopes is equal to -1. This time we will practice how to solve problems involving these lines.

Sample Problem 1: Equation of another line perpendicular to the given line

Find the equation of the line through $(1,2)$ and perpendicular to $2x-y-4=0$

Solution:

The product of the slopes of two equations is equal to -1.

Assuming that m₁ is the slope of given line and m₂ is the slope of the required line we establish the relationship below.

$m_1m_2=-1$ .

Solve for m₁:

$2x-y-4=0$

$y=2x-4$ in $y=mx+b$ ,

m is the slope the line. It is clear then that m₁=2

Solve for m₂:

$m_2 =\displaystyle\frac{-1}{m_1}$

$m_2 =\displaystyle\frac{-1}{2}$

Generate equation of the line:

Since we have the given point (1,2) and slope -1/2

$y-y_1=m_2(x-x_1)$

$y-2=\displaystyle\frac{-1}{2}(x-1)$

$2(y-2)=-1(x-1)$

$2y-4=-x+1$

$x+2y-5=0$

the latter is the required equation.

Sample problem 2: Equation of perpendicular bisector

Find the equation of perpendicular bisector of the segment joining (2,1) and (-4,7).

Solution:

Solve for slope of the segment:

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

$m=\displaystyle\frac{7-1}{-4-2}$

$m=\displaystyle\frac{6}{-6}$

$m=-1$

Treating m as m₁. It is obvious that m₂=1 for their product become -1.

Generating equation of bisector:

Since this is a bisector it will divide the segment into two equal parts and will pass through the midpoint

Let $(x_m,y_m)$ be the coordinate of the midpoint

$x_m= \displaystyle\frac{x_1+x_2}{2}$

where x₁ and x₂ are the x- coordinates of the segment given.

$x_m= \displaystyle\frac{2+(-4)}{2}$

$x_m=-1$

For y_m:

$y_m= \displaystyle\frac{y_1+y_2}{2}$

where y₁ and y₂ are the y-coordinates of the segment given

$y_m= \displaystyle\frac{1+7}{2}$

$y_m=4$

We already have the slope at the same time the point we will be able to determine the equation of the bisector right now.

$y-y_1=m_2(x-x_1)$ , y_m=y₁ and x_m=x₁

$y-4=-1(x-(-1) )$

$x+ y-3=0$

this is the required equation

Sample Problem 3: Point Reflection

The point (-1,-2) is reflected to 1^st quadrant with respect to the line x+y=0. What is the point of reflection?

Solution:

Draft the point and the line in your paper like this.

From A(-1,-2) draw a line perpendicular to x+y=0. Draft the required point. Let O is the
intersection of the lines. And d is the distance from point A to O. Then d is also the distance
from O to B.