highschool/Grade 10/Math/MPM2DZ/Unit 1: Analytical Geometry.md

# Unit 1: Analytical Geometry

- The slope of perpedicular lines are `negative reciprocal`.
- 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 $`a,b,c \in \mathbb{Z}, a>0`$ 
- `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 $`y - y_c = 0`$ 

## Triangle Centers

## Centroid
The centroid of a triangle is the common intersection of the 3 medians. The centroid is also known as the centre of mass or centre of gravity of an object (where the mass of an object is concentrated).

<img src="http://mathwords.com/c/c_assets/centroid.jpg" width="300">

### Procedure To Determine The Centroid
1. Find the equation of the two median lines. **The median is the line segment from a vertex from a vertex to the midpoint of the opposite side**.
2. Find the point of intersection using elimnation or substitution.

- Alternatively, only for checking your work, let the centroid be the point $`(x, y)`$, and the 3 other points be $`(x_1, y_1), (x_2, y_2), (x_3, y_3)`$ respectively, then the
centroid is simply at $`(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1+y_2+y_3}{3})`$

## Circumcentre
The circumcentre ($`O`$) of a triangle is the common intersection of the 3 perpendicular bisectors of the sides of a triangle.

<img src="http://mathbitsprep.org/Geometry/Constructions/perpbis1ba.jpg" width="300">

### Procedure To Determine The Centroid
1. Find the equation of the perpendicular bisectors of two sides. **A perpendicular (right) bisector is perpendicular to a side of the triangle and passes through the midpoint of that side of the triangle**.
2. Find the point of intersection of the two lines using elimination or substitution.

## Orthocentre
The orthocenter of a triangle is the common intersection of the 3 lines containing the altitudes.

<img src="https://upload.wikimedia.org/wikipedia/commons/thumb/9/93/Triangle.Orthocenter.svg/1200px-Triangle.Orthocenter.svg.png" width="300">

### Procedure To Determine The Orthocentre
1.Find the equation of two of the altitude lines. **An altitude is a perpendicular line segment from a vertex to the line of the opposite side.**
2. Find the point of intersection of the two lines using elimination or substitution.


## Classifying Shapes

<img src="https://files.catbox.moe/3cfs4h.png" width="600">


## Properties Of Quadrilaterals

<img src="https://files.catbox.moe/asixh9.png" width="500">

## Ratios
- To calculate each segment of the line given the ratio, the answer is simply 
- $`(x_1 + \dfrac{x_2 - x_1}{r}, y_1 + \dfrac{y_2 - y1}{r})`$, where $`r, (x_1,y_1) (x_2,y_2)`$ are the  **total** ratio, first point and second point respectively.
- Note that the above is for moving up a line. When moving down, we simply subtract like so:
- $`(x_2 - \dfrac{x_2 - x_1}{r}, y_2 - \dfrac{y_2 - y1}{r})`$
- For example, from a point like $`(2, 3)`$ to a point ($`5, 6)`$, and having a ratio of $`2:1`$ split at point $`P`$, the coordindates of point $`P`$ is simply 
- $`(5 - \dfrac{5-2}{3}, 7 - \dfrac{6-3}{3})`$
- Which is $`(4, 6)`$