dwrfwrw
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I done the first question but I'm not really sure how to compute Q.
I found out the magnitude of SR though https://gyazo.com/edd766f5a3e0d3f77e26e2f74844810d
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papie
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(Original post by dwrfwrw)
I done the first question but I'm not really sure how to compute Q.
I found out the magnitude of SR though https://gyazo.com/edd766f5a3e0d3f77e26e2f74844810d
What question you need help for ?
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dwrfwrw
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Second One am not really sure how to visually draw this diagram though
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dwrfwrw
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(Original post by papie)
What question you need help for ?
hello?
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mqb2766
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Have you done the distance of a point from a line formula?

If not, you could form the line and its normal which passes through the third point then use pythagoras between the point and the lines crossing point. Slightly more laborious, but not too difficult.

For drawing, just plot the points and its a good way to validate.
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dwrfwrw
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(Original post by mqb2766)
Have you done the distance of a point from a line formula?

If not, you could form the line and its normal which passes through the third point then use pythagoras between the point and the lines crossing point. Slightly more laborious, but not too difficult.

For drawing, just plot the points and its a good way to validate.
I found the distance of SR using the formula yes but I'm not really sure on what should I do next to find perpendicular distance of Q
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mqb2766
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(Original post by dwrfwrw)
I found the distance of SR using the formula yes but I'm not really sure on what should I do next to find perpendicular distance of Q
There is a formula for finding the distance of Q from the line SR if you've come across that?

If not find the equation of the line SR,
then find the equation of the line normal to it that passes through Q
Find the intersection point T of the two lines and hence use pythagoras to find the length QT
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dwrfwrw
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is QT= ROOT 5?

(Original post by mqb2766)
There is a formula for finding the distance of Q from the line SR if you've come across that?

If not find the equation of the line SR,
then find the equation of the line normal to it that passes through Q
Find the intersection point T of the two lines and hence use pythagoras to find the length QT
then for area of parallelogram= 1/2bh=1/2x SR distance x QT distance= 1/2 x root 5 x 4root5=10 unit squared thanks for helping me
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mqb2766
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(Original post by dwrfwrw)
is QT= ROOT 5?
Looks good, but upload your working if you want it checked
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dwrfwrw
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I found for part P the equation of SR and found m between (1,-2) and (9,2) and used x1 and y1 to substitute into y-y1=m(x-x1) and I got y=1/2x-5
then I found perpendicular gradient for the normal which I got m=-2x
then substituted Q(6,3) INTO Y-Y1=m(x-x1)
y-3=-2(x-6)
y-3=-2x+12
y=-2x+15(eq of normal), eq of y=1/2x-5
to find the intersection point T, I equated both eq and solved
-2x+15=1/2x-5
5/2x=35/2
x=7
sub'ed x into y=-2(7)+15 and got y=1
(7,1) is T
all I did was use distance formula between (6,3) Q and T(7,1) AND i GOT root 5
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mqb2766
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(Original post by dwrfwrw)
I found for part P the equation of SR and found m between (1,-2) and (9,2) and used x1 and y1 to substitute into y-y1=m(x-x1) and I got y=1/2x-5
then I found perpendicular gradient for the normal which I got m=-2x
then substituted Q(6,3) INTO Y-Y1=m(x-x1)
y-3=-2(x-6)
y-3=-2x+12
y=-2x+15(eq of normal), eq of y=1/2x-5
to find the intersection point T, I equated both eq and solved
-2x+15=1/2x-5
5/2x=35/2
x=7
sub'ed x into y=-2(7)+15 and got y=1
(7,1) is T
all I did was use distance formula between (6,3) Q and T(7,1) AND i GOT root 5
Looks good. Once you have SR, a quick way of checking is
https://www.wolframalpha.com/input/?...3D0.5%28x-1%29
But not much use in exams ...
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dwrfwrw
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(Original post by mqb2766)
Looks good. Once you have SR, a quick way of checking is
https://www.wolframalpha.com/input/?...3D0.5%28x-1%29
But not much use in exams ...
Thanks alot for the help, the question puzzled me for a bit though am not sure what link you showed me does though
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mqb2766
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(Original post by dwrfwrw)
Thanks alot for the help, the question puzzled me for a bit
Sure, there are a few ways to pose it, you could have said SR was a tangent to a circle of radius r^2 centered on Q where r is the minimum distance and solved for r, or you could just have used the actual formula (you may not have come across it)
https://www.intmath.com/plane-analyt...point-line.php
Or ...
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dwrfwrw
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(Original post by mqb2766)
Sure, there are a few ways to pose it, you could have said SR was a tangent to a circle of radius r^2 centered on Q where r is the minimum distance and solved for r, or you could just have used the actual formula (you may not have come across it)
https://www.intmath.com/plane-analyt...point-line.php
Or ...
never learnt the formula,sounds interesting but why does r have to be the minimum distance possible?
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mqb2766
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(Original post by dwrfwrw)
never learnt the formula,sounds interesting but why does r have to be the minimum distance possible?
If you want to find the distance of Q from a line L, all points that are the same distance r from Q form a locus (a circle). The (perpendicular) distance of Q from the line will be when the circle expands just enough so that L forms a tangent with it and the center - tangent point is perpendicular to L which is what you worked out using perpendicular intersecting lines.
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(Original post by dwrfwrw)
Thanks alot for the help, the question puzzled me for a bit though am not sure what link you showed me does though
For the link part, it calculates the distance from a point to a line, i.e. what this question is asking you to do.
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dwrfwrw
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(Original post by mqb2766)
For the link part, it calculates the distance from a point to a line, i.e. what this question is asking you to do.
Okay makes sense thx
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