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120 lines
5.3 KiB
Markdown
120 lines
5.3 KiB
Markdown
# Unit 1: Analytical Geometry
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- The slope of perpedicular lines are `negative reciprocal`.
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- The slopes of parallel lines are `the same`
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- The slope of a vertical line is `undefined`
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- The slope of a horizontal line is `0`.
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- The general equation of a line in standard form is $`ax+by+c=0`$, where $`a,b,c \in \mathbb{Z}, a>0`$
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- `Radius`: The distance from the centre of a circle to a point on the circumference of the cricle.
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- `Diameter`: the distance across a circle measured through the centre
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- `Chord`: a line segment joining two points on a curve
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- `Circle`: a set of points in the plane which are equidistant (same distance) from the centre
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## Distance Formula
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The distance between points $`A(x_1, y_1)`$ and $`B(x_2, y_2)`$ in the cartesian plane is:
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$`d = \sqrt{x^2 + y^2}`$
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$`d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}`$
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## Identifying Types of Traingles
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|Triangle|Property|
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|:-------|:-------|
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|Equilateral|3 equal sides. Each angle is 60 degrees. Can't be right angled|
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|Isoceles|2 equal sides, 2 equal angles. May be right angled|
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|Scalene|No equal sides. No equal angles. May be right angled|
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## Pythagorean Theorem Relationships
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|Formula|Statement|
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|:------|:--------|
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|$`c^2 = a^2+b^2`$|The triangle must be right angled|
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|$`c^2 < a^2 + b^2`$|The triangle is acute|
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|$`c^2 > a^2 + b^2`$|The triangle is obtuse|
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## Equation Of A Circle With Centre $`(0, 0)`$
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Let $`P(x, y)`$ be any point on the circle, and $`O`$ be the origin $`(0, 0)`$.
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Using Pythagorean Theorem,
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$`x^2+ y^2 = OP^2`$
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But, $`OP = r`$
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$`\therefore x^2 + y^2 = r^2`$ is the equation of a circle with centre $`(0, 0)`$ and radius, $`r`$.
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**Note: the coordinates of any point not on the cricle do not satisfy this equation**
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## Semi-Cricle With Radius $`r`$, And Centre $`(0, 0)`$
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If we solve for $`y`$ in the above equation $`y = \pm \sqrt{r^2-x^2}`$
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- $`y = +\sqrt{r^2-x^2}`$ is the **top half** of the circle.
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- $`y = -\sqrt{r^2-x^2}`$ is the **bottom half** of the circle
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## Equation Of A Circle With Centre $`(x, y)`$
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Let $`x_c, y_c`$ be the center
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$`(x - x_c)^2 + (y - y_c)^2 = r^2`$
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To get the center, just find a $`x, y`$ such that $`x - x_c = 0`$ and $`y - y_c = 0`$
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## Triangle Centers
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## Centroid
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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).
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<img src="http://mathwords.com/c/c_assets/centroid.jpg" width="300">
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### Procedure To Determine The Centroid
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1. Find the equation of the two median lines. **The median is the line segment from a vertex to the midpoint of the opposite side**.
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2. Find the point of intersection using elimination or substitution.
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- 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
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centroid is simply at $`(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1+y_2+y_3}{3})`$
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## Circumcentre
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The circumcentre ($`O`$) of a triangle is the common intersection of the 3 perpendicular bisectors of the sides of a triangle.
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<img src="http://mathbitsprep.org/Geometry/Constructions/perpbis1ba.jpg" width="300">
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### Procedure To Determine The Centroid
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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**.
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2. Find the point of intersection of the two lines using elimination or substitution.
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## Orthocentre
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The orthocenter of a triangle is the common intersection of the 3 lines containing the altitudes.
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<img src="https://upload.wikimedia.org/wikipedia/commons/thumb/9/93/Triangle.Orthocenter.svg/1200px-Triangle.Orthocenter.svg.png" width="300">
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### Procedure To Determine The Orthocentre
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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.**
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2. Find the point of intersection of the two lines using elimination or substitution.
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## Classifying Shapes
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<img src="https://files.catbox.moe/3cfs4h.png" width="600">
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## Properties Of Quadrilaterals
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<img src="https://files.catbox.moe/asixh9.png" width="500">
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## Ratios
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- To calculate each segment of the line given the ratio, the answer is simply
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- $`(x_1 + \dfrac{x_2 - x_1}{r}, y_1 + \dfrac{y_2 - y1}{r})`$, where $`r, (x_1,y_1) (x_2,y_2)`$ are the **total** ratio, first point and second point respectively.
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- Note that the above is for moving up a line. When moving down, we simply subtract like so:
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- $`(x_2 - \dfrac{x_2 - x_1}{r}, y_2 - \dfrac{y_2 - y1}{r})`$
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- 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
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- $`(5 - \dfrac{5-2}{3}, 7 - \dfrac{6-3}{3})`$
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- Which is $`(4, 6)`$
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## Shortest Distance From Point To a Line
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- The shortest distance is always a straightline, thus, the shortest distance from a point to a line must be **perpendicular.**
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- Thus, you can mind the slope of the line, then get the **negative reciprocal** (perpendicular slope), then find the equation of the perpendicular line.
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- After you have the 2 lines, proceed by using subsitution or elimination to find the **point of intersection**.
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- Then apply **distance formula** to find the shortest distance. |