Help with perpendicular straight line equations

[rkell48]

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]

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?)

 

[rkell48]

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

i don't understand how to do it

:/

 

[SammyS]

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 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]

If a line has equation y= mx+ b, then any line with equation y= mx+ c (same slope) is parallel to it. To be perpendicular, the second line must be of the form y= -(1/m)x+ c. That is the product of the two slopes must be -1.

 

[PeterO]

If a line has equation y= mx+ b, then any line with equation y= mx+ c (same slope) is parallel to it. To be perpendicular, the second line must be of the form y= -(1/m)x+ c. That is the product of the two slopes must be -1.

I wanted to see OP solve the parallel line to see if he/she knew a line parallel to -8y - 8 - 18x = 0 would have the formula -8y + c - 18x = 0
If OP knew to solve parallel lines that way, then he/she could simply use the fact that a line perpendicular to -8y - 8 - 18x = 0 would have the formula -18y + c + 8x = 0
In each case the given point is substituted to evaluate c.