1. ## Equation help.

The problem is:

"The points A and B have coordinates $(a,a^2)$ and $(2b,4b^2)$ respectively. Determine the gradient of AB in it's simplest form."

The equation for gradient that we use is $m=y^2-y^1 / x^2-x^1$

So this is what I've got so far

$m=y^2-y^1 / x^2-x^1$
$m=4b^2-a^2 / 2b-a$

Can someone please show me how to work through it?

Thanks.

2. Originally Posted by jumba
The problem is:

"The points A and B have coordinates $(a,a^2)$ and $(2b,4b^2)$ respectively. Determine the gradient of AB in it's simplest form."

The equation for gradient that we use is $m=y^2-y^1 / x^2-x^1$

$\textcolor{red}{m=\frac{y_2 - y_1}{x_2 - x_1}}$

So this is what I've got so far

$m=y^2-y^1 / x^2-x^1$
$m=4b^2-a^2 / 2b-a$

Can someone please show me how to work through it?

Thanks.
factor the numerator ...

$\frac{(2b-a)(2b+a)}{2b-a} = 2b+a \,\,\, ; \,\,\, a \ne 2b$

3. Use the difference of two squares formula, $A^2 - B^2 = (A -B)(A + B)$, on the top line and you can cancel the $2b - a$ term.

4. Thank-you very much skeeter and BobP. Can't believe I could notice that since teachers go on about factorising all the time.

5. Originally Posted by jumba
$m=4b^2-a^2 / 2b-a$
Careful...brackets required: (4b^2-a^2) / (2b-a)