# 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).
### 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.
### 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.
### 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
## Properties Of Quadrilaterals