From 7f92d2020645bca494b3c537271bfbbceda2ad4c Mon Sep 17 00:00:00 2001
From: James Su <amagicalsoup@gmail.com>
Date: Sun, 29 Dec 2019 18:40:01 +0000
Subject: [PATCH] Update Analytical Geometry.md

---
 .../Analytical Geometry.md                    | 175 +++++++++++++++++-
 1 file changed, 174 insertions(+), 1 deletion(-)

diff --git a/Grade 10/Math/MPM2DZ/Math Oral Presentation Questions/Analytical Geometry.md b/Grade 10/Math/MPM2DZ/Math Oral Presentation Questions/Analytical Geometry.md
index b210c3e..02a0c27 100644
--- a/Grade 10/Math/MPM2DZ/Math Oral Presentation Questions/Analytical Geometry.md	
+++ b/Grade 10/Math/MPM2DZ/Math Oral Presentation Questions/Analytical Geometry.md	
@@ -73,4 +73,177 @@ $`\therefore`$ The `orthocenter` is at $`(\dfrac{5}{4}, 2)`$
 
 ### Question 2 a)
 
-midpoint = $`\large (\dfrac{\sqrt{72} + \sqrt{32}}{2}, \dfrac{-\sqrt{12} - \sqrt{48}}{2} \large )`$
+midpoint = $` (\dfrac{\sqrt{72} + \sqrt{32}}{2}, \dfrac{-\sqrt{12} - \sqrt{48}}{2} )`$
+
+$` = ( \dfrac{6\sqrt{2} + 4\sqrt{2}}{2}, \dfrac{-2\sqrt{3}, -4\sqrt{3}}{2}) `$
+
+$` = 3\sqrt{2} + 2\sqrt{2}, -\sqrt{3 - 2\sqrt{3}`$
+
+$` = (5\sqrt{2}, -3\sqrt{3})`$
+
+$`\therefore`$ The midpoint is at $`(5 \sqrt{2}, -3\sqrt{3})`$
+
+### Question 2 b)
+
+Center of mass = centroid.
+
+Centroid = where all median lines of a trinagle intersect.
+
+$`M_{AB} = (\dfrac{8+12}{2}, \dfrac{12+4}{2}) = (10, 8)`$
+
+$`m_{M_{AB} C} = \dfrac{8-8}{10-2} = 0`$
+
+$`y_{M_{AB} C} = 8 \quad (1)`$
+
+$`M_{BC} = (\dfrac{12+2}{2}, {8+4}{2}) = (7, 6)`$
+
+$`m_{M_{BC} A} = \dfrac{6-12}{7-8} = 6`$
+
+$`y_{M_{BC}A} - 12 = 6(x-8)`$
+
+$`y_{M_{BC}A} = 6x - 48 +12`$
+
+$`y_{M_{BC}A} = 6x - 36 \quad (2)`$
+
+```math
+\begin{cases}
+
+y_{M_{BC} A} = 8 & \text{(1)} \\
+
+y_{M_{BC} A} = 6x - 36 & \text{(2)} \\
+
+\end{cases}
+```
+
+Sub $`(1)`$ into $`(2)`$
+
+$`8 = 6x - 36`$
+
+$`6x = 44`$
+
+$`x = \dfrac{44}{6} = \dfrac{22}{3} \quad (3)`$
+
+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 $`(1)`$
+
+$`y = \dfrac{-3}{4}(\dfrac{344}{25}) + 20`$
+
+$`y = \dfrac{-258}{25} + 20`$
+
+$`y = \dfrac{-257}{25} + \dfrac{500}{25}`$
+
+$`y = {242}{25}`$
+
+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.
+
+
+
+
+