Linear Equations help

[droodles]

Given the points A (a,3) B(-2,1) and C(3,2) find the possible value of if the length of AB is twice the length of BC

This is what I don't understand.

- this where I am not sure with the 2AB at the start.


....
Anyhow, down the track I ended up with a = -1 and then 12 and then 13

but the correct answer is = -12 or 8


what

[Mao]

....

[droodles]

i like that SHA BI

[Mao]

for (h,k) to lie on y=x+1 means:
(h,k)=(x,y)=(x,x+1)

distance is 5 units, therefore











or



Edit: fixed fkn n00bish mistake *slap self*

[cara.mel]

=> x=4 or x=-3
=> the answers of (4,5) and (-3,-2) you said

[Mao]

ooops... crap.... LOL

[droodles]

P and Q are the points of the intersection of the line  with the x and y axes respectively. The gradient QR is , where R is the point with the x-oordinate

Find the y-coordinate of R in terms of

I dont understand QR = 1/2 where R is the point with x-coordinate 2a

[Mao]

R is a point, and when a line is drawn through Q and R the gradient is 1/2

Q is the y-intercept of the line , rearranging equation gives us ,

if we let the point (x,y) be R, then we have:

, as you would normally calculate gradients

simplifying by cross multiplication gives us:

where

substituting gives us:





hope that helped

[droodles]

There's a triangle ABC with A(1,1) and B(-1,4). The gradients of respectively.

Find the coordinates of C.

I started off with


but for some reason I got some screwy answer because the coordinates of C are (5,7)

[cara.mel]

Someone can probably find a quicker way, but I am going to find the equation of the line AC, equation of line BC, and then find intersection

BC: y=mx+c
4 = 1/2*-1 + c
=> y = x/2 + 9/2

AC: y=mx+c
1 = 3/2 + c
=> y= 3x/2 - 1/2

Then for intersection:
x/2 + 9/2 = 3x/2 - 1/2
x + 9 = 3x - 1
2x = 10
x = 5

then sub in to either equation and y should = 7

Edit: I can't be bothered with Latex still, but you were on the right track
Not entirely sure how you tried to get an answer out of your equation, but if you had set up AC in the same way (), that pair of simultaneous equations is the same and you would have got there

[Mao]

lol u beat me to it!