# coordnate geometry

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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

I found out the magnitude of SR though https://gyazo.com/edd766f5a3e0d3f77e26e2f74844810d

Last edited by dwrfwrw; 1 week ago

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#2

(Original post by

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

**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

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Second One am not really sure how to visually draw this diagram though

Last edited by dwrfwrw; 1 week ago

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#5

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.

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.

Last edited by mqb2766; 1 week ago

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(Original post by

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.

**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.

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#7

(Original post by

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

**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

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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is QT= ROOT 5?

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

(Original post by

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

**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

Last edited by dwrfwrw; 1 week ago

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#9

(Original post by

is QT= ROOT 5?

**dwrfwrw**)is QT= ROOT 5?

Last edited by mqb2766; 1 week ago

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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

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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#11

(Original post by

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

**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

https://www.wolframalpha.com/input/?...3D0.5%28x-1%29

But not much use in exams ...

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(Original post by

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 ...

**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 ...

Last edited by dwrfwrw; 1 week ago

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#13

(Original post by

Thanks alot for the help, the question puzzled me for a bit

**dwrfwrw**)Thanks alot for the help, the question puzzled me for a bit

https://www.intmath.com/plane-analyt...point-line.php

Or ...

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(Original post by

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 ...

**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 ...

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#15

(Original post by

never learnt the formula,sounds interesting but why does r have to be the minimum distance possible?

**dwrfwrw**)never learnt the formula,sounds interesting but why does r have to be the minimum distance possible?

Last edited by mqb2766; 1 week ago

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#16

(Original post by

Thanks alot for the help, the question puzzled me for a bit though am not sure what link you showed me does though

**dwrfwrw**)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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(Original post by

For the link part, it calculates the distance from a point to a line, i.e. what this question is asking you to do.

**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.

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