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Maths A level Coordinate Geometry

Help no clue what to do for this question.

Show that the equation of SR is 2y=x-5 and find the equation of the line L through Q perpendicular to SR

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Reply 1
Avatar for davros
davros
16
Original post by DavidSmith1998
Help no clue what to do for this question.

Show that the equation of SR is 2y=x-5 and find the equation of the line L through Q perpendicular to SR


I think you're going to have to give us the full question, not just a fragment of it which doesn't mean much in isolation :smile:
Sorry :frown:.

The coordinates of four points are P(-2,-1), Q (6,3), R (9,2) and S (1,-2)

i) Calculate the gradient
ii) what is the shape

3)... part im stuck on
Reply 3
Avatar for davros
davros
16
Original post by DavidSmith1998
Sorry :frown:.

The coordinates of four points are P(-2,-1), Q (6,3), R (9,2) and S (1,-2)

i) Calculate the gradient
ii) what is the shape

3)... part im stuck on


So are you happy with finding the equation of the line SR?
Original post by davros
So are you happy with finding the equation of the line SR?


I think thats where i'm confused I'm not sure quite what to find.
Reply 5
Avatar for davros
davros
16
Original post by DavidSmith1998
I think thats where i'm confused I'm not sure quite what to find.


OK, forget the other bits of information for the moment.

Imagine you have 2 points S and R and you know their coordinates. Do you know how to find the equation of the line joining these points?
Original post by davros
OK, forget the other bits of information for the moment.

Imagine you have 2 points S and R and you know their coordinates. Do you know how to find the equation of the line joining these points?


Is it y - y1 = m(x-x1)
Reply 7
Avatar for davros
davros
16
Original post by DavidSmith1998
Is it y - y1 = m(x-x1)


That's one form of it, yes.

So can you put in your x and y values from S and R to find the gradient and then find the intercept of the line (where it crosses one of the axes)?
Original post by davros
That's one form of it, yes.

So can you put in your x and y values from S and R to find the gradient and then find the intercept of the line (where it crosses one of the axes)?


I found the gradient is -1/2 but im unsure how to find the intercept
Reply 9
Avatar for davros
davros
16
Original post by DavidSmith1998
I found the gradient is -1/2 but im unsure how to find the intercept


OK so you have y = (-1/2)x + c and you want to find c. Just plug in the x and y coordinates of one of the points S or R and you find what c is :smile:
Original post by davros
OK so you have y = (-1/2)x + c and you want to find c. Just plug in the x and y coordinates of one of the points S or R and you find what c is :smile:


ok so the y intercept is 5/2 what now? Find the perpendicular line?
Reply 11
Avatar for davros
davros
16
Original post by DavidSmith1998
ok so the y intercept is 5/2 what now? Find the perpendicular line?


I think you've gone slightly wrong somewhere - you've got y = (-1/2)x + (5/2) which is the same as 2y = -x + 5 but your original post said the answer was 2y = x - 5!

Re-check that you put the corresponding x and y values in the right place when you worked out the gradient :smile:
Original post by davros
I think you've gone slightly wrong somewhere - you've got y = (-1/2)x + (5/2) which is the same as 2y = -x + 5 but your original post said the answer was 2y = x - 5!

Re-check that you put the corresponding x and y values in the right place when you worked out the gradient :smile:


GAHH i did 2--2 = -4 :frown:(

should be 2--2 = 4
9 - 1 = 8

gradient = 1/2 instead of negative 1/2
Reply 13
Avatar for davros
davros
16
Original post by DavidSmith1998
GAHH i did 2--2 = -4 :frown:(

should be 2--2 = 4
9 - 1 = 8

gradient = 1/2 instead of negative 1/2


Great! So you've got y = (1/2)x - (5/2) (which is the same as 2y = x - 5)

Now you want another line which is perpendicular to this one. What do you know about the gradients of perpendicular lines?
Original post by davros
Great! So you've got y = (1/2)x - (5/2) (which is the same as 2y = x - 5)

Now you want another line which is perpendicular to this one. What do you know about the gradients of perpendicular lines?


They have to multiply together to = -1

So the gradient of the perpendicular line = -2
Reply 15
Avatar for davros
davros
16
Original post by DavidSmith1998
They have to multiply together to = -1

So the gradient of the perpendicular line = -2


Good. So now you know that L has equation y = -2x + k and you want to find k.

You are told that L goes through a particular point so you can put in another pair of (x,y) coordinates to find k.
Original post by davros
Good. So now you know that L has equation y = -2x + k and you want to find k.

You are told that L goes through a particular point so you can put in another pair of (x,y) coordinates to find k.


L goes through Q correct?

i subbed Q into the equation y=mx+c

3= -2(6)+c
3=-12+c
15 = C
(edited 9 years ago)
Reply 17
Avatar for davros
davros
16
Original post by DavidSmith1998
L goes through Q correct?

i subbed Q into the equation y=mx+c

3= -2(6)+c
3=-12+c
-9=c


Almost!

You are OK at 3 = -12 + c

Now what does c have to be? :biggrin:
Original post by davros
Almost!

You are OK at 3 = -12 + c

Now what does c have to be? :biggrin:


15. I edited my post earlier but obviously just missed the reply! such an idiot hahaha. What now?
Reply 19
Avatar for davros
davros
16
Original post by DavidSmith1998
15. I edited my post earlier but obviously just missed the reply! such an idiot hahaha. What now?


I think you're done aren't you? You have found the equation of L.

I need sleep now :smile:

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