2 hard maths questions (well for me anyways)?

Question

where does the line 3y-2x + 9 = 0 cut the y-axis and what is its gradient?

Find the equation of the line through T(-2,6) which is perpendicular tot heline through A(1,-2) and B(4,4)

Can everyone please leave their working so i can work it out for next time. Thank You

PS sorry if i keep posting up maths questions its just that i never really got it in my class and i really want to pass my exam this year!

Answer 1

For the first part, put the eq in slope intercept form, or basically, solve for y.

3y=2x-9

y = 2/3x -3

so -3 is the y intercept.

Srry 'bout the second part, I don't do gradients...

Answer 2

The way to tackle these problems is to re-arrange to the form:

y = m*x + c

where:

- m = the gradient

- c = the y-intercept

(a) Therefore:

3y - 2x + 9 = 0

Re-arranges to:

y = (2/3) * x - 3

Therefore:

- m = gradient = 2/3

- c = y intercept = -3

(b) We firstly need to find the gradient of the line we are perpendicular to, so using y = m*x + c

(A) -2 = m * 1 + c

(B) 4 = m * 4 + c

So we have:

m + c = -2

4m + c = 4

Solving by subtracting these two equations:

3m = 6 --> m = 2

You always determine the gradient of a line perpendicular as:

= - 1/m

Therefore, we want the gradient of the new line to be = -1/2

So:

y = c - x / 2

We know it need to go through (-2,6), so we can calculate c:

6 = c + 1  --> c = 5

So the equation of the new line is:

y = 5 - x / 2

Or:

2y + x = 10