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Update Unit 1: Analytical Geometry.md

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## Analytical Geometry
## Analytical Geometry Part 1
### Question 1 a)
@ -126,124 +126,3 @@ $`x = \dfrac{44}{6} = \dfrac{22}{3} \quad (3)`$
By $`(1)`$, $`y=8`$.
$`\therefore`$ The centroid is at $`(\dfrac{22}{3}, 8)`$
### Question 3
Shortest distance = straight perpendicular line that connets $`A`$ to a point on line $`\overline{GH}`$
$`M_{GH} = \dfrac{42+30}{38 + 16} = \dfrac{72}{54} = \dfrac{4}{3}`$
$`M_{\perp GH} = \dfrac{-3}{4}`$
$`y_{\perp GH} - 32 = \dfrac{-3}{4}(x+16)`$
$`y_{\perp GH} = \dfrac{-3}{4}x + 20 \quad (1)`$
$`y_{GH} + 30 = \dfrac{4}{3}(x+16)`$
$`y_{GH} = \dfrac{4}{3}x - \dfrac{26}{3} \quad (2)`$
```math
\begin{cases}
y_{\perp GH} = \dfrac{-3}{4}x + 20 & \text{(1)} \\
\\
y_{GH} = \dfrac{4}{3}x - \dfrac{26}{3} & \text{(2}) \\
\end{cases}
```
Sub $`(1)`$ into $`(2)`$
$`\dfrac{-3}{4}x + 20 = \dfrac{4}{3}x - \dfrac{26}{3}`$
$`-9x + (12)20 = 16x - 4(26)`$
$`25x = 344`$
$`x = \dfrac{344}{25} \quad (3)`$
Sub $`(3)`$ into $`(1)`$
$`y = \dfrac{-3}{4}(\dfrac{344}{25}) + 20`$
$`y = \dfrac{-258}{25} + 20`$
$`y = \dfrac{-258}{25} + \dfrac{500}{25}`$
$`y = \dfrac{242}{25}`$
Distance $`= \sqrt{(-16-\dfrac{344}{25})^2 + (32 - \dfrac{242}{25})^2} = 37.2`$
$`\therefore`$ The shortest length pipe is $`37.2`$ units.
### Question 4
Let $`(x, y)`$ be the center of the circle, and $`r`$ be the radius of the circle.
```math
\begin{cases}
(x-4)^2 + (y-8)^2 = r^2 & \text{(1)} \\
(x-5)^2 + (y-1)^2 = r^2 & \text{(2)} \\
(x+2)^2 + y^2 = r^2 & \text{(3)} \\
\end{cases}
```
Sub $`(1)`$ into $`(2)`$
$`x^2 - 8x + 16 + y^2 - 16y + 64 = x^2 - 10x + 25 + y^2 -2y + 1`$
$`-8x -16y + 80 = -10x - 2y + 26`$
$`2x - 14y = -54`$
$`x - 7y = -27 \quad (4)`$
Sub $`(2)`$ into $`(3)`$
$`x^2 - 10x + 25 + y^2 - 2y + 1 = x^2 + 4x + 4 + y^2`$
$`-10x - 2y +26 = 4x + 4`$
$`14x + 2y = 22`$
$`7x + y = 11`$
$`y = 11 - 7x \quad (5)`$
Sub $`(5)`$ into $`(4)`$
$`x - 7(11-7x) = -27`$
$`x - 77+ 49x = 27`$
$`50x = 50`$
$`x = 1 \quad (6)`$
Sub $`(6)`$ into $`(5)`$
$`y = 11 - 7(1)`$
$`y = 4 \quad (7)`$
Sub $`(6), (7)`$ into $`(3)`$
$`(1+2)^2 + 4^2 = r^2`$
$`r^2 = 16 + 9`$
$`r^2 = 25`$
$`\therefore (x-1)^2 + (y-4)^2 = 25`$ is the equation of the circle.