# Perpendicular Lines and its Application

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

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

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

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

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

$2y-4=-x+1$

$x+2y-5=0$

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 ym:

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

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.

$y-y_1=m_2(x-x_1)$    , ym=y1 and xm=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 1st 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.

Solve the equation of the line through (-1,-2) and perpendicular to x+y=0

Slope of given equation is -1. So the slope of perpendicular line is 1.

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

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

$x-y-1=0$

Solve for intersection of the line:

x+y=0       – eqn.1

x-y-1=0    – eqn.2

Solving the equations simultaneously the system is (1/2, -1/2)

The intersection or point O is the midpoint of the line.

From the midpoint formula,

Solve for x1:

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

$\frac{1}{2}= \displaystyle\frac{x_1+(-1)}{2}$

$\frac{1}{2}(2)= x_1+(-1)$

$x_1=2$

Solve for y1:

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

$\frac{-1}{2}=\displaystyle\frac{y_1+(-2)}{2}$

$\frac{-1}{2}(2)= y_1+(-2)$

$y_1=1$

The required point is (2,1).

Practice Problems:

1. Find the equation of the line through (0,1) and perpendicular to 3x-2y=1.

2. Find the equation of the line through (2,-7) and perpendicular to y-axis.

3. Find the equation of perpendicular bisector of the (3,2) and (-1,6).

4. The point (3,2) is reflected with respect to y-axis. What is the point of reflection?

5. The point (2,-4) is reflected with respect to equation 3x+y-3=0. What is the point of reflection?

Answer key will be available soon.