From 7fa388a3b330a0a3e6de75002087bac1883ee016 Mon Sep 17 00:00:00 2001
From: James Su <amagicalsoup@gmail.com>
Date: Sun, 12 Jan 2020 23:54:32 +0000
Subject: [PATCH] Update Unit 1: Analytical Geometry.md

---
 ... => Unit 1: Analytical Geometry Part 1.md} | 123 +-----------------
 1 file changed, 1 insertion(+), 122 deletions(-)
 rename Grade 10/Math/MPM2DZ/Math Oral Presentation Questions/{Unit 1: Analytical Geometry.md => Unit 1: Analytical Geometry Part 1.md} (57%)

diff --git a/Grade 10/Math/MPM2DZ/Math Oral Presentation Questions/Unit 1: Analytical Geometry.md b/Grade 10/Math/MPM2DZ/Math Oral Presentation Questions/Unit 1: Analytical Geometry Part 1.md
similarity index 57%
rename from Grade 10/Math/MPM2DZ/Math Oral Presentation Questions/Unit 1: Analytical Geometry.md
rename to Grade 10/Math/MPM2DZ/Math Oral Presentation Questions/Unit 1: Analytical Geometry Part 1.md
index 411b6b1..62de48b 100644
--- a/Grade 10/Math/MPM2DZ/Math Oral Presentation Questions/Unit 1: Analytical Geometry.md	
+++ b/Grade 10/Math/MPM2DZ/Math Oral Presentation Questions/Unit 1: Analytical Geometry Part 1.md	
@@ -1,4 +1,4 @@
-## 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.
-
-
-
-
-