C1 question

Q.
Line m with equation 2x + 3y = 15 meets l at Point C (l equation is y = 3/2x - 3/2)

Determine the coordinates of the point C

y = 3/2x - 3/2 [1] /// 2x + 3y = 15 [2]

Sort [2] out

y = -2/3x + 5

3/2x - 3/2 = -2/3x + 5 [put into groups]

3/2x + 2/3x = 5 + 3/2
(x2)(x3) [to remove fractions]

9x + 4x = 30 + 9
13x = 39
x = 39/13

2(39/13) + 3y = 15

78/13 + 3y = 15
(-78/13)

3y = 195/13 - 78/13
3y = 117//113 (/3)
y = 39/113

(39/13 , 39/113) but it says the correct answer is (3,3) what have I done wrong?? Been confused for the last 10 minutes.

Original post by ckfeister

...

Why are you sorting [2] out? Just substitute [1] into [2] for y. Much less mess that way.

Original post by ckfeister

...

Also 39/13 is 3. You are failing to manipulate fractions beyond that point.

Reply 3

A_A123

1

You've made a small error in the last paragraph where you have changed the denominator from 13 to 113:

Instead of:
3y = 195/13 - 78/13
3y = 117//113 (/3)
y = 39/113

You get:
3y = 195/13 - 78/13
3y = 117//13 (/3)
y = 39/13 = 3

(edited 7 years ago)

Reply 4

Jack1066

17

Revising for C1 resit already?

Reply 5

ckfeister

OP

19

Original post by A_A123

You've made a small error in the last paragraph where you have changed the denominator from 13 to 113:

Instead of:
3y = 195/13 - 78/13
3y = 117//113 (/3)
y = 39/113

You get:
3y = 195/13 - 78/13
3y = 117//13 (/3)
y = 39/13 = 3

Oops... here I'm mixed up too.... I keep getting mixed numbers (I don't know the gradient, I tried using AB gradient as ABC is a right angel but I get 11 not 6... and the original one I get k = 9)

A(-1,-2)
B(7,2)
C(k, 4)
Gradient of AB is 1/2 (if thats anything)

Calculate the value of k

(x, y)
B(7,2) [1]
C(k,4) [2]

m = y2 - y1 / x2 - x1
4 - 2 / k - 7 = m
(x k-7)

2 = k - 7
k = 9 but it says k = 6 ???

Reply 6

ckfeister

OP

19

Original post by Jack1066

Revising for C1 resit already?

Haven't even done first AS yet..?

Original post by ckfeister

Oops... here I'm mixed up too.... I keep getting mixed numbers (I don't know the gradient, I tried using AB gradient as ABC is a right angel but I get 11 not 6... and the original one I get k = 9)

A(-1,-2)
B(7,2)
C(k, 4)
Gradient of AB is 1/2 (if thats anything)

Calculate the value of k

(x, y)
B(7,2) [1]
C(k,4) [2]

m = y2 - y1 / x2 - x1
4 - 2 / k - 7 = m
(x k-7)

2 = k - 7
k = 9 but it says k = 6 ???

If ABC is a right angle then gradient of CB will be the negative reciprocal of the gradient AB. No need to use any line equations, simply the change in y over change in x. Draw a diagram, it might help.

C1 question

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you