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 and perpendicular to
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.
Solve for m1:
m is the slope the line. It is clear then that m1=2
Solve for m2:
Generate equation of the line:
Since we have the given point (1,2) and slope -1/2
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).
Solve for slope of the segment:
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 be the coordinate of the midpoint
where x1 and x2 are the x- coordinates of the segment given.
where y1 and y2 are the y-coordinates of the segment given
We already have the slope at the same time the point we will be able to determine the equation of the bisector right now.
, ym=y1 and xm=x1
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?
Draft the point and the line in your paper like this.
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.
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:
Solve for y1:
The required point is (2,1).
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.