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# Coordinate Geometry watch

1. The vertices of a triangle are A(2,4), B(-8,2) and C(4,-6). The points P and Q are the midpoint of AB and AC respectively. The line through P perpendicular to AB meets the line through Q perpendicular to AC at R.
1. Show that the equation of PR is y+5x+12=0 and find the equation of QR.
2. Find the co-ordinates of R.
3. Show that BR=CR

So these are the working I've got, and I'm a little confused.
Midpoint AB =
Midpoint AB = P
P =
Midpoint AC =
Midpoint AC = Q
Q =
Perpendicular to AB =
So this means PR = So is this okay for a?
2. (Original post by Abusive)
The vertices of a triangle are A(2,4), B(-8,2) and C(4,-6). The points P and Q are the midpoint of AB and AC respectively. The line through P perpendicular to AB meets the line through Q perpendicular to AC at R.
1. Show that the equation of PR is y+5x+12=0 and find the equation of QR.
2. Find the co-ordinates of R.
3. Show that BR=CR

So these are the working I've got, and I'm a little confused.
Midpoint AB =
Midpoint AB = P
P =
Midpoint AC =
Midpoint AC = Q
Q =
Perpendicular to AB =
So this means PR =
Now how do I answer the questions?
You have your coordinates of P correct. You almost got Q - it should be which is what I think you meant.

Next you need the gradient AB which you've worked out incorrectly - it should be - and you want the perpendicular gradient to AB after you find this. Construct the equation through P, perpendicular to AB, as shown in the question. This is your line .

Next do the same thing at Q. You know it's coordinate, now you need the gradient of AC and get the perpendicular gradient to it before constructing which is perpendicular to AC through Q.

is where and intersect, I assume you know how to find R??

The for part C it's just applying Pythagoras' Theorem to show distances are equal.
3. (Original post by RDKGames)
You have your coordinates of P correct. You almost got Q - it should be which is what I think you meant.

Next you need the gradient AB which you've worked out incorrectly - it should be - and you want the perpendicular gradient to AB after you find this. Construct the equation through P, perpendicular to AB, as shown in the question. This is your line .

Next do the same thing at Q. You know it's coordinate, now you need the gradient of AC and get the perpendicular gradient to it before constructing which is perpendicular to AC through Q.

is where and intersect, I assume you know how to find R??

The for part C it's just applying Pythagoras' Theorem to show distances are equal.
Hey, I've updated the original post. Would that suffice as an answer for a?
Also yes I know how to do the rest I'm sure. Thanks for your help!
4. (Original post by Abusive)
Hey, I've updated the original post. Would that suffice as an answer for a?
Also yes I know how to do the rest I'm sure. Thanks for your help!
It's hard to read as it stands but if you're essentially taking and substituting with before solving for , then you're fine. Except I'm not sure why you're saying " so " as that is not needed and doesn't show anything.

Once you derive that's all there is to it.
5. (Original post by RDKGames)
It's hard to read as it stands but if you're essentially taking and substituting with before solving for , then you're fine. Except I'm not sure why you're saying " so " as that is not needed and doesn't show anything.

Once you derive that's all there is to it.
Thanks for your help means a lot.

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