Maths help (2 Viewers)

[qwerty44]

Alright, another problem.

I need to find the equation of a straight line passing through (0, -2) and perpendicular to the line x - 2y = 9. What I have tried to do so far is get the gradient of the line which is 1/2 and from there, I'm not sure what to do.

Help please

Find the gradient of the line x-2y=9.

Now because it has to be perpendicular, the negative reciprocal of that will be the gradient you use for the straight line. i.e You found the gradient to be 1/2, but to be perpendicular, you take the negative reciprocal which is -2. That's the gradient for the new line.

Then just use y-y₁=m(x-x₁) with m=-2 and the points given.

 

[Carrotsticks]

Alright, another problem.

I need to find the equation of a straight line passing through (0, -2) and perpendicular to the line x - 2y = 9. What I have tried to do so far is get the gradient of the line which is 1/2 and from there, I'm not sure what to do.

Help please

1. The gradient of that line given is 1/2, but we are given that the line we want to find is perpendicular to it. This means it has gradient -2.

2. You already have the point.

3. Point/Gradient formula.

 

[Shazer2]

Thanks heaps qwerty44, I'll try and get the rest done now.

 

[Shazer2]

Quick question, to get the gradient of the perpendicular, is there a formula or how do I do it?

 

[qwerty44]

The perpendicular to any gradient is the negative reciprocal, there is no formula, its just that. You can test by observation from a graphing site ( www.fooplot.com ) so you understand it more.

e.g if a gradient of a line is -3, then the gradient of the line perpendicular to it is 1/3. See here what it looks like.

It is like that with every gradient.

 

[Shazer2]

Alright, can I just do 1/gradient? Anyway, here is my working. I got the wrong answer, and I'm not sure where I go wrong.

 

[qwerty44]

Alright, can I just do 1/gradient? Anyway, here is my working. I got the wrong answer, and I'm not sure where I go wrong.

You just did the reciprocal, not the negative reciprocal. m should = -1/(-3/2)=2/3 not -2/3.

You can't just do 1/gradient, must be -1/gradient.

 

[Shazer2]

Thanks heaps qwerty44, you're a life saver.

 

[qwerty44]

Thanks heaps qwerty44, you're a life saver.

No worries

It's good that your getting help, but try as hard as you can to find the problem before asking. Look through notes and worked example to see how it's done and if after that you can't see your mistake, then post.

 

[Shazer2]

They throw in the odd question that I don't understand, yet they don't provide a worked example for it. I'm stuck on this one now: "Find the equation of the perpendicular bisector of the line passing through (6, -3) and (-2, 1)." I got the gradient (1/2) from the 2 points, but I don't know what to do next.

 

[Carrotsticks]

They throw in the odd question that I don't understand, yet they don't provide a worked example for it. I'm stuck on this one now: "Find the equation of the perpendicular bisector of the line passing through (6, -3) and (-2, 1)." I got the gradient (1/2) from the 2 points, but I don't know what to do next.

We have the gradient of the line, but we want the perpendicular gradient, which is the negative reciprocal. So the gradient of the line we want is -2.

Now we need a point (in order to use the Point/Gradient formula). We are given that the point is the bisector (midpoint) of the two given points. Find the midpoint.

Now use the formula.

 

[qwerty44]

They throw in the odd question that I don't understand, yet they don't provide a worked example for it. I'm stuck on this one now: "Find the equation of the perpendicular bisector of the line passing through (6, -3) and (-2, 1)." I got the gradient (1/2) from the 2 points, but I don't know what to do next.

Perpendicular bisector means the line perpendicular to the line passing through those points, but also goes through the middle of those two points. So find the midpoint of the two points [ (2,?) ] and then use the point gradient formula. (are you sure the gradient is 1/2?)

 

[qwerty44]

We have the gradient of the line, but we want the perpendicular gradient, which is the negative reciprocal. So the gradient of the line we want is -2.

Now we need a point (in order to use the Point/Gradient formula). We are given that the point is the bisector (midpoint) of the two given points. Find the midpoint.

Now use the formula.

Isn't it 2?

 

[Carrotsticks]

Isn't it 2?

He said the gradient of the line is 1/2, so the perpendicular gradient is the negative reciprocal ie: -2.

Unless he was mistaken.

 

[Shazer2]

Thanks CarrotSticks and qwerty44, I got that question. Now for the next one. Stay tuned

My apologies, it should be -1/2.

 

[Wealth]

r u retarded?

These questions are just so easy.

 

[Shazer2]

Yeah, and? Might be easy for you, but not so much me. I'm trying to get better.

 

[Carrotsticks]

Yeah, and? Might be easy for you, but not so much me. I'm trying to get better.

Nevermind him, he is a known troll.

 

[deswa1]

Yeah, and? Might be easy for you, but not so much me. I'm trying to get better.

This is good, just keep practicing. What I would do (I'm not sure if you do this now) is try and keep your working really organised so you know all the information you have and need. For example, for the above equation where you wanted the perp. bisector, you need both the gradient of that line and a point that it passes through.

Structure your working like:
Gradient of original line is X.
Therefore gradient of perp. bisector is -1/X
Perp. bisector passes through the midpoint.
The midpoint is (A,B).
Therefore the equation of the perpendicular bisector is y-B=(-1/X)(x-A)

When you get more used to it, you won't need to do all this but for now it should help- also try and draw a diagram if you can to help visualise.

 

[Shazer2]

Nevermind him, he is a known troll.

It's all good, at least I'm getting somewhere.