From fa3b3d3b3d2f689a65b81bb96f76bef6b6ae3adb Mon Sep 17 00:00:00 2001
From: James Su <amagicalsoup@gmail.com>
Date: Tue, 17 Sep 2019 20:59:59 +0000
Subject: [PATCH] Update Unit 1: Analytical Geometry.md

---
 .../MPM2DZ/Unit 1: Analytical Geometry.md     | 58 ++++++++++++++++++-
 1 file changed, 57 insertions(+), 1 deletion(-)

diff --git a/Grade 10/Math/MPM2DZ/Unit 1: Analytical Geometry.md b/Grade 10/Math/MPM2DZ/Unit 1: Analytical Geometry.md
index f14f5ea..279fa68 100644
--- a/Grade 10/Math/MPM2DZ/Unit 1: Analytical Geometry.md	
+++ b/Grade 10/Math/MPM2DZ/Unit 1: Analytical Geometry.md	
@@ -4,4 +4,60 @@
 - The slopes of parallel lines are `the same`
 - The slope of a vertical line is `undefined`
 - The slope of a horizontal line is `0`.
-The general equation of a line in standard form is $`ax+by+c=0`$, where $`a,b,c \in \mathbb{Z}, a>0`$ 
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+- The general equation of a line in standard form is $`ax+by+c=0`$, where $`a,b,c \in \mathbb{Z}, a>0`$ 
+- `Radius`: The distance from the centre of a circle to a point on the circumference of the cricle.
+- `Diameter`: the distance across a circle measured through the centre
+- `Chord`: a line segment joining two points on a curve
+- `Circle`: a set of points in the plane which are equidistant (same distance) from the centre
+
+## Distance Formula
+
+The distance between points $`A(x_1, y_1)`$ and $`B(x_2, y_2)`$ in the cartesian plane is:
+
+$`d = \sqrt{x^2 + y^2}`$
+
+$`d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}`$
+
+## Identifying Types of Traingles
+
+|Triangle|Property|
+|:-------|:-------|
+|Equilateral|3 equal sides. Each angle is 60 degrees. Can't be right angled|
+|Isoceles|2 equal sides, 2 equal angles. May be right angled|
+|Scalene|No equal sides. No equal angles. May be right angled|
+
+## Pythagorean Theorem Relationships
+
+|Formula|Statement|
+|:------|:--------|
+|$`c^2 = a^2+b^2`$|The triangle must be right angled|
+|$`c^2 < a^2 + b^2`$|The triangle is acute|
+|$`c^2 > a^2 + b^2`$|The triangle is obtuse|
+
+## Equation Of A Circle With Centre $`(0, 0)`$
+
+Let $`P(x, y)`$ be any point on the circle, and $`O`$ be the origin $`(0, 0)`$.
+
+Using Pythagorean Theorem,
+
+$`x^2+ y^2 = OP^2`$
+
+But, $`OP = r`$
+
+$`\therefore x^2 + y^2 = r^2`$ is the equation of a circle with centre $`(0, 0)`$ and radius, $`r`$.
+
+**Note: the coordinates of any point not on the cricle do not satisfy this equation**
+
+## Semi-Cricle With Radius $`r`$, And Centre $`(0, 0)`$
+
+If we solve for $`y`$ in the above equation $`y = \pm \sqrt{r^2-x^2}`$
+- $`y = +\sqrt{r^2-x^2}`$ is the **top half** of the circle.
+- $`y = -\sqrt{r^2-x^2}`$ is the **bottom half** of the circle
+
+## Equation Of A Circle With Centre $`(x, y)`$
+
+Let $`x_c, y_c`$ be the center
+
+$`(x - x_c)^2 + (y - y_c)^2 = r^2`$ 
+
+To get the center, just find a $`x, y`$ such that $`x - x_c = 0`$ and $`y - y_c = 0`$ 
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