Can somebody please solve this Past Exam C1 question?

Can somebody please help me to solve this past exam question for me. There are 3 parts to the question and I have done the first part, but cannot remember how to complete the second and third part.
3. The line L1 passes through the point (9,-4) and has gradient 1/3
A. Find an equation for L1 in the form ax+by+c=0, where a, b and c are integers.
Answer= x-3y-21=0
B. The line L2 passes through the Origin O and has gradient -2. The lines L1 and L2 intersect at the point P. Calculate the coordinates of P.
C. Given that L1 crosses the y-axis at the point C. Calculate the exact area of (picture of a triangle) OCP.

If somebody could solve this and show your working so I can get this drummed in my head, I would be very grateful.

Reply 1

Olibert

11

B) Put the equations of L1 and L2 equal to each other, cancel out etc etc etc...

Reply 2

AveUgotaWkdSide

OP

But what is the equation for L2?

Reply 3

DeanK2

The line passses through the origin and has gradient -2.

y=-2x

Reply 4

AveUgotaWkdSide

OP

Nope still don't get it, can somebody help me a bit further please

Reply 5

Ragzy

AveUgotaWkdSide

Can somebody please help me to solve this past exam question for me. There are 3 parts to the question and I have done the first part, but cannot remember how to complete the second and third part.
3. The line L1 passes through the point (9,-4) and has gradient 1/3
A. Find an equation for L1 in the form ax+by+c=0, where a, b and c are integers.
Answer= x-3y-21=0
B. The line L2 passes through the Origin O and has gradient -2. The lines L1 and L2 intersect at the point P. Calculate the coordinates of P.
C. Given that L1 crosses the y-axis at the point C. Calculate the exact area of (picture of a triangle) OCP.

If somebody could solve this and show your working so I can get this drummed in my head, I would be very grateful.

b) here we are given L2: m= -2 @ (0,0)

So equation of L2: y-0 = -2(x-0)
y=-2x

Therefore to find P, you do simultaneous eqations and equate.

L1: 3y = x - 21
L2: y= -2x ---------> 3y= -6x

therfore equate L1 and L2: -6x = x -21
7x = 21
x = 3, (now substitute back into the original equations of L1 or L2 -----> y= -2 * 3 = -6

Therefore P: (3,-6)

c) We know L1: 3y= x - 21
And we are told it crosses the y axis, which ofc implies x=0

Therefore we can find point c: if x=0, using the equation of L1:3y= x - 21

We find that 3y = -21

therefore y=-7

So C: (0, -7)

So we have O: (0,0) C: (0,-7) and P: (3,-6)

Now you can break this triangle down into two further triangles (see pic attached)

work out the area of each one... using half base*height rule

and you get: Total area = (0.5 *3*6) + (0.5*1*3)
= 9+1.5
= 10.5 (units^2)

I hope thats helped!!

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Reply 6

crshbr

that's the same thing...

edit to be helpful:

(x/3)-7=-2x

rearrange this to get a value for x, then put the x value into the first equation to get a value for y. this will be the point where they intersect.

edit again:

the (x/3)-7 comes from the first equation. I just rearranged it into the form y=... ( i hope that's clear)