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Help with C1 - Coordinate Geometry
I've posted this on the A level forum but I think I will get more help/response here..
Aha, I've been revising for a while.. but I just came across a question..
In coordinate geometry, i've been given 4 coordinates. Asked to draw it and show, by calculation, that it is a trapezium. Similar questions say show, by calculation, that it is a parallelogram etc.
Coordinates
A=0,0
B=0,3
C=6,6
D=12,6
Anyone know how I can do this? Thanks for the help!
Aha, I've been revising for a while.. but I just came across a question..
In coordinate geometry, i've been given 4 coordinates. Asked to draw it and show, by calculation, that it is a trapezium. Similar questions say show, by calculation, that it is a parallelogram etc.
Coordinates
A=0,0
B=0,3
C=6,6
D=12,6
Anyone know how I can do this? Thanks for the help!
Ok, a trapesium has one pair of parallel lines and one pair that are not. In parallel lines you have the same gradient.
So if you draw sketch using the co-ordinates (or you can tell this from looking at the question) you will get the idea that BC should be parallel to AD, and BA should NOT be parallel to CD.
Gradient = Y2-Y1/X2-X1
BC= 3-6/0-6 = -3/-6 = 1/2
AD= 0-6/0-12 = -6/-12 = 1/2
Here we have proved that BC and AD are parallel as that are both y=1/2x + c
BA= 3-0/0-0 = 3
CD= 6-6/6-12 = -6/-6 = 1
Here we have proved that BA and CD are not parallel which is what I was expecting.
So, as far as I know that is the question answered.
If the shape was a parrallelogram, you would expect to find both the sets of opposite lines to have the same gradient.
Hope I helped
So if you draw sketch using the co-ordinates (or you can tell this from looking at the question) you will get the idea that BC should be parallel to AD, and BA should NOT be parallel to CD.
Gradient = Y2-Y1/X2-X1
BC= 3-6/0-6 = -3/-6 = 1/2
AD= 0-6/0-12 = -6/-12 = 1/2
Here we have proved that BC and AD are parallel as that are both y=1/2x + c
BA= 3-0/0-0 = 3
CD= 6-6/6-12 = -6/-6 = 1
Here we have proved that BA and CD are not parallel which is what I was expecting.
So, as far as I know that is the question answered.
If the shape was a parrallelogram, you would expect to find both the sets of opposite lines to have the same gradient.
Hope I helped
emilylikeeee
Ok, a trapesium has one pair of parallel lines and one pair that are not. In parallel lines you have the same gradient.
So if you draw sketch using the co-ordinates (or you can tell this from looking at the question) you will get the idea that BC should be parallel to AD, and BA should NOT be parallel to CD.
Gradient = Y2-Y1/X2-X1
BC= 3-6/0-6 = -3/-6 = 1/2
AD= 0-6/0-12 = -6/-12 = 1/2
Here we have proved that BC and AD are parallel as that are both y=1/2x + c
BA= 3-0/0-0 = 3
CD= 6-6/6-12 = -6/-6 = 1
Here we have proved that BA and CD are not parallel which is what I was expecting.
So, as far as I know that is the question answered.
If the shape was a parrallelogram, you would expect to find both the sets of opposite lines to have the same gradient.
Hope I helped
So if you draw sketch using the co-ordinates (or you can tell this from looking at the question) you will get the idea that BC should be parallel to AD, and BA should NOT be parallel to CD.
Gradient = Y2-Y1/X2-X1
BC= 3-6/0-6 = -3/-6 = 1/2
AD= 0-6/0-12 = -6/-12 = 1/2
Here we have proved that BC and AD are parallel as that are both y=1/2x + c
BA= 3-0/0-0 = 3
CD= 6-6/6-12 = -6/-6 = 1
Here we have proved that BA and CD are not parallel which is what I was expecting.
So, as far as I know that is the question answered.
If the shape was a parrallelogram, you would expect to find both the sets of opposite lines to have the same gradient.
Hope I helped
That really helped! Thank you SO much
pluuum
That really helped! Thank you SO much
Not a problem! It helps me to help other people any way. Good luck with your exams in June
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