# Equation of the line perpendicular to.....

• Feb 13th 2010, 07:24 AM
Beginner
Equation of the line perpendicular to.....
Hi

I am back studying after a few years away from it and I wonder if someone will be able to give a bit of advice.
The question i am needing help with is this-
Determine the equation of the line perpendicular to 4x+3Y+5=0 passing through the point (2,4).
The notes I have been given for the Maths class are very poor and was wondering if someone could help or maybe send me a link to a good tutorial website.

Thanks
• Feb 13th 2010, 08:13 AM
HallsofIvy
Quote:

Originally Posted by Beginner
Hi

I am back studying after a few years away from it and I wonder if someone will be able to give a bit of advice.
The question i am needing help with is this-
Determine the equation of the line perpendicular to 4x+3Y+5=0 passing through the point (2,4).
The notes I have been given for the Maths class are very poor and was wondering if someone could help or maybe send me a link to a good tutorial website.

Thanks

Any (non-vertical) line can be written in the "slope-intercept" form y= mx+ b where m is the "slope" and b is the "y-intercept". Write your given line in that form to find its slope. A line perpendicular to y= mx+ b has slope -1/m. You can use the fact that it passes through (2, 4) to find b.
• Feb 13th 2010, 08:36 AM
teuthid
In order to write the equation for a line we need two pieces of information:
(1) A point on the line
(2) The slope of the line

This problem gives us a point for free: (2,4). Unfortunately we're going to have to work to find the slope. We know that our line is going to be perpendicular to their line (4x+3y+5=0), so if we can find the slope of their line we'll know what ours should be (I'll explain how in a little bit).

Finding the Slope of their line:
======================
To find the slope of a line from its equation just solve for y, and whatever the x coefficient turns out to be, that is the slope...

$\begin{array}{lrcl}
&4x+3y+5 &=& 0\\
\Rightarrow& 4x+3y &=&-5\\
\Rightarrow& 3y&=&-4x-5\\
\Rightarrow& y &=& -\frac{4}{3}x-\frac{5}{3}
\end{array}$

So the slope of their line is $-\frac{4}{3}$

Find the Slope of our line:
===================
If they had asked us to find a parallel line, we would have our slope already--we'd just use the slope of their line. However, since we're looking for a perpendicular line, we have to make two adjustments to their slope to get ours:
(1) change the sign
(2) take the reciprocal

Changing the sign and taking the reciprocal of $-\frac{4}{3}$ gives us $\frac{3}{4}$.

Writing the equation:
================
When you've got a point (let's call it $(x_0,y_0)$) and a slope (let's call it $m$), the easiest equation to write is point-slope form:

$y-y_0=m(x-x_0)$

For us, $(x_0,y_0)$ is (2,4) and $m$ is $\frac{3}{4}$, so just plug the appropriate values in and you'll have your equation!