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# Coordinate geometry straight line q watch

1. The triangle PQR has the vertices P(8,6), Q(0,2) and R(2,r). Find the value of r when the triangle has a right angle at P.

I'm confused, does this mean that lines PQ and PR are perpendicular? By doing this method & solving it, I get the r value of -2. The answer is 18. How?
2. (Original post by Magenta96)
The triangle PQR has the vertices P(8,6), R(0,2) and R(2,r). Find the value of r when the triangle has a right angle at P.

I'm confused, does this mean that lines PQ and PR are perpendicular? By doing this method & solving it, I get the r value of -2. The answer is 18. How?
Yes it does mean they are perpendicular

I assume (0,2) is actually Q

18 is correct
3. (Original post by TenOfThem)
Yes it does mean they are perpendicular

I assume (0,2) is actually Q

18 is correct
oh okay, managed to get the answer of 18 after making silly arithmetic errors the first time round
4. Ok. I'm not sure how good I will be at explaining this but here goes...

Q(0,2) and P, and PR meet at P. Since point P is a right angle, PQ and PR are perpendicular. Tell help, you can draw a circle with a right angled triangle inside it.

Ok so if P is a right angle that means the gradient of PQ and PR multiply to give -1.

The gradient of a line is y2-y1/x2-x1 ( / means fraction)

The gradient of PR is therefore: R-6/2-8 so R-6/-6

The gradient of PQ is therefore 6-2/8-0. this simplifies to give us 1/2.

Ok now all we have to do is find R.

R-6/-6 x 1/2 = -1 since we know the gradients are going to multiply to give us -1

After multiplying the fractions you should be left with R-6=12
R is therefore equal to 18.

I hope this helped!
5. (Original post by JackersAFC)
Ok. I'm not sure how good I will be at explaining this but here goes...

Q(0,2) and P, and PR meet at P. Since point P is a right angle, PQ and PR are perpendicular. Tell help, you can draw a circle with a right angled triangle inside it.

Ok so if P is a right angle that means the gradient of PQ and PR multiply to give -1.

The gradient of a line is y2-y1/x2-x1 ( / means fraction)

The gradient of PR is therefore: R-6/2-8 so R-6/-6

The gradient of PQ is therefore 6-2/8-0. this simplifies to give us 1/2.

Ok now all we have to do is find R.

R-6/-6 x 1/2 = -1 since we know the gradients are going to multiply to give us -1

After multiplying the fractions you should be left with R-6=12
R is therefore equal to 18.

I hope this helped!
omg i thank you so much

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