bambi_eyes
Badges: 4
Rep:
?
#1
Report Thread starter 9 years ago
#1
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
0
reply
lukas1051
Badges: 18
Rep:
?
#2
Report 9 years ago
#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.
0
reply
Tulian
Badges: 0
Rep:
?
#3
Report 9 years ago
#3
(Original post by 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 .
0
reply
bambi_eyes
Badges: 4
Rep:
?
#4
Report Thread starter 9 years ago
#4
(Original post by 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 .
I tried that but I got r=10 and thats wrong
0
reply
Gemini92
Badges: 0
Rep:
?
#5
Report 9 years ago
#5
(Original post by 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
If there is a right angle at point P, then the lines PQ and PR are perpendicular to each other.
0
reply
bambi_eyes
Badges: 4
Rep:
?
#6
Report Thread starter 9 years ago
#6
(Original post by Gemini92)
If there is a right angle at point P, then the lines PQ and PR are perpendicular to each other.
Thank you!!!!
0
reply
taha_17
Badges: 1
Rep:
?
#7
Report 5 years ago
#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.
1
reply
davros
  • Study Helper
Badges: 16
Rep:
?
#8
Report 5 years ago
#8
(Original post by 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
0
reply
Sahil kamboj
Badges: 0
Rep:
?
#9
Report 3 years ago
#9
Could anyone answer the (iv) partIs isosceles with RQ=RP.....PLEASE
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Feeling behind at school/college? What is the best thing your teachers could to help you catch up?

Extra compulsory independent learning activities (eg, homework tasks) (5)
4.46%
Run extra compulsory lessons or workshops (18)
16.07%
Focus on making the normal lesson time with them as high quality as possible (21)
18.75%
Focus on making the normal learning resources as high quality/accessible as possible (14)
12.5%
Provide extra optional activities, lessons and/or workshops (35)
31.25%
Assess students, decide who needs extra support and focus on these students (19)
16.96%

Watched Threads

View All