# A level Maths help

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#1
An engineer is making a circular disc.
A model of the disc is drawn as a circle on a set of axes such that it passes
through coordinate points P (–1, –1), Q (5, k) and R(7, 3)
Given that angle PRQ is 90°, find the value of k and the area of the disc

I can’t figure out how to do this I drew a diagram and I don’t see a right angle could someone please show me how to do it
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1 month ago
#2
Anything about semicircles and right angles ring any bells?
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1 month ago
#3
(Original post by Anonymous9795)
An engineer is making a circular disc.
A model of the disc is drawn as a circle on a set of axes such that it passes
through coordinate points P (–1, –1), Q (5, k) and R(7, 3)
Given that angle PRQ is 90°, find the value of k and the area of the disc

I can’t figure out how to do this I drew a diagram and I don’t see a right angle could someone please show me how to do it
If PRQ is 90 degrees, what is the scalar product of PR and RQ?
0
1 month ago
#4
(Original post by Anonymous9795)
An engineer is making a circular disc.
A model of the disc is drawn as a circle on a set of axes such that it passes
through coordinate points P (–1, –1), Q (5, k) and R(7, 3)
Given that angle PRQ is 90°, find the value of k and the area of the disc

I can’t figure out how to do this I drew a diagram and I don’t see a right angle could someone please show me how to do it
are you sure it is PRQ that is a right angle??
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#5
(Original post by DFranklin)
Anything about semicircles and right angles ring any bells?
Nope I’m a first year so haven’t done anything in a while the only thing I managed to come up with is (5,7) but that’s where I get confused about where the circle is on the graph I assumed that r was the midpoint and p was on the circumference but that would mean that (5,7) would just be along the diameter.
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#6
(Original post by df1)
are you sure it is PRQ that is a right angle??
Yeah that’s what it says
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1 month ago
#7
(Original post by Anonymous9795)
Nope I’m a first year so haven’t done anything in a while the only thing I managed to come up with is (5,7) but that’s where I get confused about where the circle is on the graph I assumed that r was the midpoint and p was on the circumference but that would mean that (5,7) would just be along the diameter.
Agree with (5,7) as the position of Q. All of P,Q,R are on the circumference.

Since PRQ is a right angle, it follows that PQ is a diameter. Hence work out radius and area.
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1 month ago
#8
(Original post by Anonymous9795)
An engineer is making a circular disc.
A model of the disc is drawn as a circle on a set of axes such that it passes
through coordinate points P (–1, –1), Q (5, k) and R(7, 3)
Given that angle PRQ is 90°, find the value of k and the area of the disc

I can’t figure out how to do this I drew a diagram and I don’t see a right angle could someone please show me how to do it
I am uploading a link to my worked solution, this is how I would've done it, sorry the sketch is not the best . https://imgur.com/crl9ZeI
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#9
(Original post by ghostwalker)
Agree with (5,7) as the position of Q. All of P,Q,R are on the circumference.

Since PRQ is a right angle, it follows that PQ is a diameter. Hence work out radius and area.
Thanks for your help I got confused with where the circle was but now l understand
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1 month ago
#10
(Original post by Anonymous9795)
Thanks for your help I got confused with where the circle was but now l understand
@/https://imgur.com/crl9ZeI This is my worked solution and how I would do the question ignore the @/ and just copy and paste the link after ittt
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#11

Thanks but I don’t think that q can be (5,10.75) because it wouldn’t be on the line perpendicular (y=-2x+17) to PR (y=0.5x+0.5) thus making PRQ not 90°
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1 month ago
#12
(Original post by Anonymous9795)

Thanks but I don’t think that q can be (5,10.75) because it wouldn’t be on the line perpendicular (y=-2x+17) to PR (y=0.5x+0.5) thus making PRQ not 90°
yeah i guess that would make my answer wrong then, I can't think of another way to do it but I might have made a numerical or simplifying error somewhere so you could always retry it and be more careful than me and see if it gets an answer that works lol
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1 month ago
#13
(Original post by df1)
yeah i guess that would make my answer wrong then, I can't think of another way to do it but I might have made a numerical or simplifying error somewhere so you could always retry it and be more careful than me and see if it gets an answer that works lol
actually you could try a method of simultaneous equations of the lines PR and QR to solve for K
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1 month ago
#14
(Original post by Anonymous9795)
Thanks but I don’t think that q can be (5,10.75) because it wouldn’t be on the line perpendicular (y=-2x+17) to PR (y=0.5x+0.5) thus making PRQ not 90°
I think it was agreed above that Q is the point (5, 7) which you had already worked out for yourself.

So now you know the coordinates of all the points you can work out the diameter, and hence the radius, and hence the area as suggested earlier 1
#15
(Original post by df1)
actually you could try a method of simultaneous equations of the lines PR and QR to solve for K
I just substituted x=5 in to y=-2x+17 which gave me (5,7)
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