# Help with perpendicular straight line equations

## Homework Statement

Find the equation of the line perpendicular to

## Homework Equations

−8y − 8 − 16x = 0 and passing through the point (18, 0).

## The Attempt at a Solution

-8y = 16x +8
y = =2x -1
-2(x-18) = y - 0
-2x +36 = 0
-2 x = -36

NascentOxygen
Staff Emeritus
You are given the equation of a line. Re-arrange it into the familiar form: y=mx + c

From this, you can immediately identify the gradient of that line.

Next step, determine what the gradient would be of any line that is perpendicular to this one.

You would have covered this in class; or at least, you'll be able to find it in your textbook: if a line has a gradient m, what is the gradient of a line perpendicular to this? (Maybe you can pretend that you don't actually know, but can show how it can be worked out from first principles?)

my lecturer is awful, my lecture notes are brief and don't explain anything

i don't understand how to do it

:/

SammyS
Staff Emeritus
Homework Helper
Gold Member

## Homework Statement

Find the equation of the line perpendicular to

## Homework Equations

−8y − 8 − 16x = 0 and passing through the point (18, 0).

## The Attempt at a Solution

-8y = 16x +8
y = =2x -1
-2(x-18) = y - 0
-2x +36 = 0
Why did y disappear?
-2 x = -36

PeterO
Homework Helper

## Homework Statement

Find the equation of the line perpendicular to

## Homework Equations

−8y − 8 − 16x = 0 and passing through the point (18, 0).

## The Attempt at a Solution

-8y = 16x +8
y = =2x -1
-2(x-18) = y - 0
-2x +36 = 0
-2 x = -36

You need to be familiar with the gradient relationship for perpendicular lines

Even before you can use that, perhaps you could find the equation of the line parallel to

−8y − 8 − 16x = 0 and passing through the point (18, 0).

so we can assess you ability to do problems of this type.

Parallel is often considered the simpler problem, with perpendicular just adding one extra twist.

HallsofIvy