Equation of a straight line given one point and perpendicular to a line.
In this tutorial we continue with the straight line lessons in this time knowing one point and perpendicular to a line.
Today we continue with our tutorials the Equations of a straight line, If you haven’t read them I invite you to do it in the following links.
Equation of a straight line given one point and slope.
Equation of a straight line given two points.
As always we are going to find the equation following this 4 steps:
- Find the slope.
- Use point slope equation.(( Y-Y1)=m(X-X1))
- Solve for Y.
- Plot the equation.
Example:
Given the point (1,2) and the equation must be perpendicular to the line 1X+2Y=3
- Find the slope.
In this case we don’t have any information about the slope of the equation so first we will have to find the slope of the perpendicular line to get some information.
For this we will find Y of the perpendicular line.
1X+2Y=3
2Y=3–1X
Y=(3–1X)/2
Taking in mind the equation of the line (Y=mX+c), we know now the slope is the constant attached to the X so the slope of the perpendicular line is:
m=(-1/2)
But to find the slope of our equation we have to find the negative reciprocal of the slope of the given perpendicular line.
for doing that we change the top and the bottom in of m and multiply it by -1.
m=(-1/2)
mpp=-1(-2/1)
mpp=2
mpp is going to be the slope of our straight line so now that we found it we can continue to the next step.
2.Use point slope equation.(( Y-Y1)=m(X-X1))
As we have (1,2) as our given point we know that X1=1 and Y1=2 and mpp=2 so we are going to remplace them in the equation.
( Y-Y1)=mpp(X-X1)
(Y-2)=2(X-1)
3.Solve for Y.
(Y-2)=2(X-1)
(Y-2)=2X-2
Y=2X-2+2
Y=2X
4.Plot the equation
Now with the help of Quick Graph I decided to show you the two perpendicular lines.
In these case we can see
Y=2X in red
and
1X+2Y=3 in green
If you want to practice your plots or learn more about graphing an equation use Quick Graph.
If you already have Quick Graph, press here to see this equation.
