# AS Maths-Help needed!!

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So I'm really stuck on the following question:

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

I asked my teacher and he said lines Q and R are perpendicular so their gradients multiply to give -1 but I don't think you can find the gradient of a line through one co ordinate.

I know r=18 but I still want to to know why

Please help, plus rep for those who do

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

I asked my teacher and he said lines Q and R are perpendicular so their gradients multiply to give -1 but I don't think you can find the gradient of a line through one co ordinate.

I know r=18 but I still want to to know why

Please help, plus rep for those who do

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#2

Try drawing it, it will help. You can plot two of the points, you can draw two perpendicular lines, and you can draw a range on which the point you don't yet know will lie.

Use pythagoras, trigonometry as appropriate.

Use pythagoras, trigonometry as appropriate.

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#3

(Original post by

So I'm really stuck on the following question:

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

I asked my teacher and he said lines Q and R are perpendicular so their gradients multiply to give -1 but I don't think you can find the gradient of a line through one co ordinate.

I know r=18 but I still want to to know why

Please help, plus rep for those who do

**bambi_eyes**)So I'm really stuck on the following question:

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

I asked my teacher and he said lines Q and R are perpendicular so their gradients multiply to give -1 but I don't think you can find the gradient of a line through one co ordinate.

I know r=18 but I still want to to know why

Please help, plus rep for those who do

Okay so to start with, there is going to be a right angle when the gradient is equal to -1 . So use the gradient formula between two points, the point P and the point R .

Gradient is change in y/change in x, change in y will be (6-r)/(8-2) set it to -1 .

Try rearranging that to get r .

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(Original post by

Okay so to start with, there is going to be a right angle when the gradient is equal to -1 . So use the gradient formula between two points, the point P and the point R .

Gradient is change in y/change in x, change in y will be (6-r)/(8-2) set it to -1 .

Try rearranging that to get r .

**Tulian**)Okay so to start with, there is going to be a right angle when the gradient is equal to -1 . So use the gradient formula between two points, the point P and the point R .

Gradient is change in y/change in x, change in y will be (6-r)/(8-2) set it to -1 .

Try rearranging that to get r .

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#5

**bambi_eyes**)

So I'm really stuck on the following question:

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

I asked my teacher and he said lines Q and R are perpendicular so their gradients multiply to give -1 but I don't think you can find the gradient of a line through one co ordinate.

I know r=18 but I still want to to know why

Please help, plus rep for those who do

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(Original post by

If there is a right angle at point P, then the lines PQ and PR are perpendicular to each other.

**Gemini92**)If there is a right angle at point P, then the lines PQ and PR are perpendicular to each other.

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#7

Correct answers for part 1 2 3 P(8,6) Q(0,2) R(2,r),

We have to find the value of "r" when the triangle

Part 1: hasa right angle at P

First, m1m2=-1

so, PQ= -1/PR

PQ= 6-2/8-0 = 1/2

PR= 6-r/8-2 = 6-r/6

Then, 1/2 = -1/(6-r/6) multiply 6 by -1 which will give 1/2 = -6/6-r

Cross multiply: 6-r = -12

6+12=r

r = 18.

Part 2: has a right angle at Q

First, QP = 2-6/0-8 = 1/2

QR = 2-r/0-2 = 2-r/-2

Then,

1/2 = -1/(2-r/-2) multiply -2 by -1

which will give 1/2 = 2/2-r

Cross multiply = 2-r = 4

2-4 = r and r= -2

Part 3: has a right angle at R

First, RQ = r-6/2-8 = r-6/-6

RQ = r-2/2-0 = r-2/2

Then,

m1m2= -1

r-6/-6 = -1/(r-2/2)

r-6/-6 = -2/r-2

r-2 (r-6) = -6 (-2)

r^2 -6r -2r +12 = 12

r^2 -8r = 0

r (r-8) = 0

r = 0 or r-8=0

r = 8

In part 4 it says,

Find the value of r when the triangle is isosceles with RQ = RP. I tried to solve but couldn't make it.

Hope this answer would be helpful to you.

Taha Faruqi.

We have to find the value of "r" when the triangle

Part 1: hasa right angle at P

First, m1m2=-1

so, PQ= -1/PR

PQ= 6-2/8-0 = 1/2

PR= 6-r/8-2 = 6-r/6

Then, 1/2 = -1/(6-r/6) multiply 6 by -1 which will give 1/2 = -6/6-r

Cross multiply: 6-r = -12

6+12=r

r = 18.

Part 2: has a right angle at Q

First, QP = 2-6/0-8 = 1/2

QR = 2-r/0-2 = 2-r/-2

Then,

1/2 = -1/(2-r/-2) multiply -2 by -1

which will give 1/2 = 2/2-r

Cross multiply = 2-r = 4

2-4 = r and r= -2

Part 3: has a right angle at R

First, RQ = r-6/2-8 = r-6/-6

RQ = r-2/2-0 = r-2/2

Then,

m1m2= -1

r-6/-6 = -1/(r-2/2)

r-6/-6 = -2/r-2

r-2 (r-6) = -6 (-2)

r^2 -6r -2r +12 = 12

r^2 -8r = 0

r (r-8) = 0

r = 0 or r-8=0

r = 8

In part 4 it says,

Find the value of r when the triangle is isosceles with RQ = RP. I tried to solve but couldn't make it.

Hope this answer would be helpful to you.

Taha Faruqi.

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#8

(Original post by

Correct answers for part 1 2 3

....

Hope this answer would be helpful to you.

Taha Faruqi.

**taha_17**)Correct answers for part 1 2 3

....

Hope this answer would be helpful to you.

Taha Faruqi.

Hi, welcome to TSR.

Please don't post full solutions - it's against Forum guidelines!

And please don't resurrect four-year-old threads.

Thanks

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