1. ## coordinate geometry.....

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

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

All working appreciated

2. 1.

You can easily see that x coordinates of B and A are equal. Thus we know that A is (a,something). This is enough because all points on the line satisfies this equation: $\displaystyle 3x+2y=24$. We know $\displaystyle x=a$, solve the equation for y.

2.
$\displaystyle y=3x^2$
$\displaystyle y=12x-12$

$\displaystyle 3x^2=12x-12$

$\displaystyle x^2-4x+4 = 0$

$\displaystyle (x-2)^2=0$

$\displaystyle x=2$

It means two graphs intersect only at x=2. There only one intersection. But, when a parabola and a line intersects at only one point, it's more like.. touching. Take a look at the graph and you'll easily see what I mean:

So, the line is tangent to the parabola.

3. 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.

4. My interpretation of the perpendicular part is that the perpendicular drawn is perpendicular to 3x+2y=24, ie, gradient of -(1/3), am I correct?

5. Originally Posted by Bartimaeus
My interpretation of the perpendicular part is that the perpendicular drawn is perpendicular to 3x+2y=24, ie, gradient of -(1/3), am I correct?
Ah, I probably misunderstood that. See Reckoner's solution, he got it correct.

6. Originally Posted by Reckoner

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

$\displaystyle \Rightarrow \frac{3(8 - a)(8 - a)}{13} = 27\Rightarrow(8-a)^2 = 117$.
okay, how did you go from $\displaystyle \Rightarrow \frac{3(8 - a)(8 - a)}{13} = 27$ to $\displaystyle \Rightarrow(8-a)^2 = 117$?
where did the 24 go? and the - in front of the 3a from th previous line? and why (8-a)(8-a)?
It doesn't quite add up to me...please explain further and add in any working you may have left out

sorry for any inconvenience

7. Originally Posted by Bartimaeus
okay, how did you go from $\displaystyle \Rightarrow \frac{3(8 - a)(8 - a)}{13} = 27$ to $\displaystyle \Rightarrow(8-a)^2 = 117$?
where did the 24 go? and the - in front of the 3a from th previous line? and why (8-a)(8-a)?
It doesn't quite add up to me...please explain further and add in any working you may have left out

sorry for any inconvenience

$\displaystyle 24 - 3a = 3(8 - a)$

8. Originally Posted by Bartimaeus
okay, how did you go from $\displaystyle \Rightarrow \frac{3(8 - a)(8 - a)}{13} = 27$ to $\displaystyle \Rightarrow(8-a)^2 = 117$?
where did the 24 go? and the - in front of the 3a from th previous line? and why (8-a)(8-a)?
It doesn't quite add up to me...please explain further and add in any working you may have left out

sorry for any inconvenience
No problem. I'll start back at the beginning of that one. Did you get the same answer that I got for $\displaystyle A$, $\displaystyle \left(\frac{4a+72}{13},\;\frac{48-6a}{13}\right)$? Anyway, we have

$\displaystyle \text{\color{red}Area }=\frac12{\color{blue}b}{\color{magenta}h}$ (Formula for area of a triangle)

$\displaystyle \Rightarrow{\color{red}27} = \frac12{\color{blue}(8 - a)}{\color{magenta}\frac{48 - 6a}{13}}$ (Substitution)

$\displaystyle \Rightarrow\frac{(48 - 6a)(8 - a)}{26} = 27$ (Simplifying)

$\displaystyle \Rightarrow\frac{6(8 - a)(8 - a)}{26} = 27$ (Factoring: $\displaystyle 48 - 6a = 6(8 - a)$)

$\displaystyle \Rightarrow\frac{3(8 - a)(8 - a)}{13} = 27$ (Reducing fraction)

$\displaystyle \Rightarrow\frac{3(8 - a)^2}{13} = 27$ (Rewriting with exponent)

$\displaystyle \Rightarrow(8 - a)^2 = \frac{27\cdot13}3 = 117$ (Multiplying through by $\displaystyle \frac{13}3$)

$\displaystyle \Rightarrow8 - a = \pm\sqrt{117}\Rightarrow a = 8\pm\sqrt{117}$

Either solution should work (it depends on which side of the $\displaystyle x$-axis A is on; everything works out the same either way). Is that clearer?

9. Originally Posted by Reckoner
b = -\frac23a[/tex]

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

10. Originally Posted by Bartimaeus
how do you find the y value? of A that is
Do you understand how I found the equation of the perpendicular line? If so, we have two lines intersecting at $\displaystyle A$:

$\displaystyle 3x + 2y = 24$

$\displaystyle y = \frac23(x - a)$

All you need to do is find their intersection, which involves solving these two equations simultaneously. Rewriting the second equation produces

$\displaystyle 3y = 2(x - a)\Rightarrow2x - 3y = 2a$

So now we have

$\displaystyle \left\{\begin{array}{rcrcrl} 3x & + & 2y & = & 24 & \quad\text{\color{red}(1)}\\ 2x & - & 3y & = & 2a & \quad\text{\color{red}(2)} \end{array}\right.$

Multiply (1) by 2 and multiply (2) by 3:

$\displaystyle \left\{\begin{array}{rcrcrl} 6x & + & 4y & = & 48 & \quad\text{\color{red}(3)}\\ 6x & - & 9y & = & 6a & \quad\text{\color{red}(4)} \end{array}\right.$

Now subtract (4) from (3):

$\displaystyle 0x + 13y = 48 - 6a\Rightarrow y = \frac{48 - 6a}{13}$

This is our value for $\displaystyle y$. To get $\displaystyle x$, we back-substitute:

$\displaystyle 3x + 2y = 24$

$\displaystyle \Rightarrow3x + 2\left(\frac{48-6a}{13}\right) = 24$

$\displaystyle \Rightarrow3x + \frac{12(8-a)}{13} = 24$ (Factoring)

$\displaystyle \Rightarrow x = 8 - \frac{4(8 - a)}{13}$ (Dividing by 3 and subtracting the fraction)

$\displaystyle \Rightarrow x = \frac{104 - 4(8 - a)}{13}$ (Combining fractions)

$\displaystyle \Rightarrow x = \frac{4a + 72}{13}$ (Expanding and simplifying)

Thus, the coordinates of $\displaystyle A$ are $\displaystyle \left(\frac{4a + 72}{13},\;\frac{48 - 6a}{13}\right)$, which agrees with the answer I gave earlier.