coordnate geometry

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

(edited 2 years ago)

Reply 1

papie

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 ?

Reply 2

dwrfwrw

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

(edited 2 years ago)

Reply 3

dwrfwrw

Original post by papie

What question you need help for ?

hello?

Reply 4

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.

(edited 2 years ago)

Reply 5

dwrfwrw

Original post by mqb2766

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

Reply 6

mqb2766

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

Reply 7

dwrfwrw

is QT= ROOT 5?

Original post by mqb2766

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

(edited 2 years ago)

Reply 8

mqb2766

Original post by dwrfwrw

is QT= ROOT 5?

Looks good, but upload your working if you want it checked

(edited 2 years ago)

Reply 9

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

Reply 10

mqb2766

Original post by dwrfwrw

Looks good. Once you have SR, a quick way of checking is
https://www.wolframalpha.com/input/?i=distance+of+%286%2C3%29+from+%28y%2B2%29%3D0.5%28x-1%29
But not much use in exams ...

Reply 11

dwrfwrw

Original post by mqb2766

Looks good. Once you have SR, a quick way of checking is
https://www.wolframalpha.com/input/?i=distance+of+%286%2C3%29+from+%28y%2B2%29%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

(edited 2 years ago)

Reply 12

mqb2766

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-analytic-geometry/perpendicular-distance-point-line.php
Or ...

Reply 13

dwrfwrw

Original post by mqb2766

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

Reply 14

mqb2766

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.