Originally Posted by

**Bartimaeus** 1. Point A lies on the line 3x+2y=24. A perpendicular is drawn from A to the x-axis, meeting at B

a) if B=(a,0) find the coordinates of A in terms of a

Check out this crude drawing:

Code:

\ /
\ /
A\/
/\
/ \ 3x + 2y = 24
/ \
-o------\-----
B=(a, 0) \

We know the perpendicular must be of the form $\displaystyle y = \frac23x + b$ (since the two slopes must multiply to -1). Since the $\displaystyle x$-intercept is at $\displaystyle (a,\;0)$, we have

$\displaystyle 0 = \frac23a + b\Rightarrow b = -\frac23a$

So the perpendicular line is $\displaystyle y = \frac23(x - a)$. The intersection of these two lines will be $\displaystyle A$: find it.

Originally Posted by

**Bartimaeus** b) The triangle bounded by AB, 3x+2y=24 and the x-axis has an area of 27 square units. Find coordinates of A

Here's a picture:

Code:

\
o A
/|\
/ | \
/ | \
/ h| \ 3x + 2y = 24
/ | \
/ | \ (8,0)
-o-------------o--
B b \

You should have gotten $\displaystyle \left(\frac{4a+72}{13},\;\frac{48-6a}{13}\right)$ for $\displaystyle A$. Thus we have

$\displaystyle \text{Area }=\frac12bh = \frac{24-3a}{13}\left(8 - a\right) = 27$

$\displaystyle \Rightarrow \frac{3(8 - a)(8 - a)}{13} = 27\Rightarrow(8-a)^2 = 117$.

You should be able to take it from here.

Originally Posted by

**Bartimaeus** 2.a) Solve equations y=3x^2 and y=12x-12 simulatneously

b) explain why y=12x-12 is a tangent to parabola y=3x^2

wingless has this covered.

I realize that I may have misinterpreted question 1, but I'll leave my work as it is since I'm not sure.