#1

May 2006 C1

The line L1 passes through the points P(-1,2) and Q(11,8).

(a) Find an equation for L1 in the form y=mx+c, where m and c are constants

The line L2 passes through the point R(10,0) and is perpendicular to L1. The lines L1 and L2 intersect at the point S.

(b) Calculate the coordinates of S.

(c)Show that the length of RS is 3(5)^0.5 This is 3 multiply the square root of 5.

(d) Hence, or otherwise, find the exact area of triangle PQR.

I've managed parts (a) and (c) but I could do with help with the other two.

Thank you very much.

The line L1 passes through the points P(-1,2) and Q(11,8).

(a) Find an equation for L1 in the form y=mx+c, where m and c are constants

The line L2 passes through the point R(10,0) and is perpendicular to L1. The lines L1 and L2 intersect at the point S.

(b) Calculate the coordinates of S.

(c)Show that the length of RS is 3(5)^0.5 This is 3 multiply the square root of 5.

(d) Hence, or otherwise, find the exact area of triangle PQR.

I've managed parts (a) and (c) but I could do with help with the other two.

Thank you very much.

#2

for part b)

m of line 2 = -1/m of line 1

then use (y-y1)= m(x-x1)

EDIT: if that made ANY sense at all.

drawing a diagram for d might help?

m of line 2 = -1/m of line 1

then use (y-y1)= m(x-x1)

EDIT: if that made ANY sense at all.

drawing a diagram for d might help?

#3

to get the perpendicular gradient all you need to do is flip it and negatize it eg if the gradient is 3 the perp grad would be -1/3

so get the value m and then make it -1/m then you've got the one value of X and Y (10,0) so put that in and get the equation of the line.

Then to get the point S put the equations equal to each other and the values for x and y that you get will be the point S.

cant figure out (d) off the top of my head cos as soon as i finished my A-levels i forgot most of it.

Thats off the top of my head

so get the value m and then make it -1/m then you've got the one value of X and Y (10,0) so put that in and get the equation of the line.

Then to get the point S put the equations equal to each other and the values for x and y that you get will be the point S.

cant figure out (d) off the top of my head cos as soon as i finished my A-levels i forgot most of it.

Thats off the top of my head

#4

I'm on it. Post back shortly.

#5

so glad i didnt do futher maths

:O

WTF

:O

WTF

#6

use the line L1 as the base of the triangle and use the distance RS as the perp height.

so (half value for L1)* 3(5)^0.5

I think

so (half value for L1)* 3(5)^0.5

I think

#7

Part B: once you have the two equations. use simulataneos equations. The results are the co-ordinates. Part D: i believe yes you draw a diagram and use pythagoras to find the area.

Yeh i guess ure exams on Wed - im shittin it 2 man. Gd luk !!

Yeh i guess ure exams on Wed - im shittin it 2 man. Gd luk !!

#8

**b)**Well, assuming for a) you got:

y = 0.5x + 2.5

The gradient for the perpendicular will be:

-1/0.5 = -2

because the gradients of two perpendicular line must equal -1 (-2 x 0.5 = -1).

As L2 goes through (10,0), this must be true:

y = -2x + c

0 = -2 x 10 + c

c = 20

So, the new equation for L2 is: y = -2x + 20.

From a): y = 0.5x + 2.5, so:

-2x + 20 = 0.5 + 2.5

2.5x = 17.5

x = 7

y = -2 x 7 + 20

y = 6

Answer: (7,6)

EDIT: I'll just start on D and post it in a few minutes.

EDIT 2: I hope it made sense.

#9

I just did the May 2006 paper which this on eis on and got between 55 and 60 out of the 75 marks. I"m just rubbish at it.. Thanks for everyone's efforts and good luck to you too slaveaway

#10

Yup, I got the exam on Wednesday too. I haven't started revising yet...

#11

You seem good at it though

#12

Thanks , it's chemistry I'm worried about lol, I'm liking maths at the moment.

For this I will use ./ for a square root sign.

3./5 is the height of the triange. The two sides (not the base) are 12 and 6, worked out from the coordinates. 12^2 is 144 and 6^2 is 36, added together is 180.

./180 = ./9./20 = 3./20 = 3./4./5 = 3 x 2./5 = 6./5 is the length of the base.

Area of triange = 1/2 base x height.

1/2 x 6./5 x 3./5 = 3./5 x 3./5 = 45

EDIT: Again, I hope this makes sense. Tell me if you don't follow anything. I'm usually very bad at explaining things.

For this I will use ./ for a square root sign.

**d)**If I were you I'd draw a diagram, it helps loads.3./5 is the height of the triange. The two sides (not the base) are 12 and 6, worked out from the coordinates. 12^2 is 144 and 6^2 is 36, added together is 180.

./180 = ./9./20 = 3./20 = 3./4./5 = 3 x 2./5 = 6./5 is the length of the base.

Area of triange = 1/2 base x height.

1/2 x 6./5 x 3./5 = 3./5 x 3./5 = 45

EDIT: Again, I hope this makes sense. Tell me if you don't follow anything. I'm usually very bad at explaining things.

#13

Right, for part (b), the first thing you need to do is work out an equation for L2, this is fairly simple. As L2 is perpendicular to L1, the gradient of L2 is found by flipping over the gradient of L1 and inverting the sign. Hence 1/2 becomes -2. Next, put this, as well as the information about which point it passes through, into:

y - y1 = m(x - x1)

giving you:

y = -2x + 20

Then you can set the equations of L1 and L2 equal to each other (0.5x + 2.5 = -2x + 20), work that through and you will get x = 7. Put this x value into either of the line equations and you get y = 6, and there is your answer (7,6).

For part (d), start by working out the length of PQ, using the coordinates of P and Q, we can form a right angled triangle of side length: PQ, 12 and 6, with PQ being the hypotenuse. Hence the legnth of PQ is the sqaure root of 144 + 36, or 6(5)^. Put this into the formula for the area of a triangle 1/2(b)(h), along with the length of RS for your height and you will get the answer of 9(5)^.

So far as I know, that should be right, good luck with the exam, I assume your taking it this coming Wednesday?

EDIT: Answer to (d) is 9 * 5, or 45, heh, went a little too quickly there...

y - y1 = m(x - x1)

giving you:

y = -2x + 20

Then you can set the equations of L1 and L2 equal to each other (0.5x + 2.5 = -2x + 20), work that through and you will get x = 7. Put this x value into either of the line equations and you get y = 6, and there is your answer (7,6).

For part (d), start by working out the length of PQ, using the coordinates of P and Q, we can form a right angled triangle of side length: PQ, 12 and 6, with PQ being the hypotenuse. Hence the legnth of PQ is the sqaure root of 144 + 36, or 6(5)^. Put this into the formula for the area of a triangle 1/2(b)(h), along with the length of RS for your height and you will get the answer of 9(5)^.

So far as I know, that should be right, good luck with the exam, I assume your taking it this coming Wednesday?

EDIT: Answer to (d) is 9 * 5, or 45, heh, went a little too quickly there...

#14

i think the answer to d is 45

L1 between p and Q = (12^2 + 6^2)^.5 = (144+36)^.5 = 180^.5 =(9^0.5)*(20^0.5)=

(3*(4^0.5)*(5^0.5))=3*2*5^.5=6(5^0.5)

L1=6 root 5

half of L1= 3 root 5

(3 root 5) Squared =9*5=45

If that makes sense

L1 between p and Q = (12^2 + 6^2)^.5 = (144+36)^.5 = 180^.5 =(9^0.5)*(20^0.5)=

(3*(4^0.5)*(5^0.5))=3*2*5^.5=6(5^0.5)

L1=6 root 5

half of L1= 3 root 5

(3 root 5) Squared =9*5=45

If that makes sense

#15

i remember this stuff!

i failed AS maths last year

i failed AS maths last year

#16

^ For d) did you get 3./5 x 3./5? Because that isn't 9./5.

3 x ./5 x 3 x ./5 = 3 x 3 x ./5 x ./5 = 9 x 5 = 45.

I think... correct me if I'm wrong.

Ah you corrected yourself, nevermind. I made the same mistake at first

3 x ./5 x 3 x ./5 = 3 x 3 x ./5 x ./5 = 9 x 5 = 45.

I think... correct me if I'm wrong.

Ah you corrected yourself, nevermind. I made the same mistake at first

#17

Thanks very much

You've helped me out a lot.

You've helped me out a lot.

#18

The answer for (d) is 45

#19

Are you doing further maths as well? (And therefore taking C1, C2 and M1 this coming week)

#20

No problem . And I'm just doing normal maths. Kinda wish I'd taken further maths though instead of English lit, which I dropped.

Aw, what did you get? If I'm gonna fail something this year it will probably be music, which is a depressing thought

i remember this stuff!

i failed AS maths last year

Aw, what did you get? If I'm gonna fail something this year it will probably be music, which is a depressing thought

#21

I got an A for maths AS but came out wit a C overall, I think I got an E average in A2

#22

Its a worthwhile a-level, just a lot of work and pressure.

#23

I'll get a C for my Maths AS.

Then two As and a B for subjects Economics,History and Politics. Dunno which one will get the B...

Then two As and a B for subjects Economics,History and Politics. Dunno which one will get the B...

#24

I've got this on wednesday too, not looking forward to it. I'll be happy with a B. Expecting a C.

#25

No problem . And I'm just doing normal maths. Kinda wish I'd taken further maths though instead of English lit, which I dropped.

Aw, what did you get? If I'm gonna fail something this year it will probably be music, which is a depressing thought

i got a U! but it was because i only attended 37% of lessons, which is a bad idea, really.

i got an A for gcse...and that was with only doing 2/3 of the coursework and one of the exams (i was ill most of the year...). so yeah. the moral of the story is, AS levels suck, but sadly to pass them one must attend lessons

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