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).

Procedure To Determine The Centroid

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.
2. Find the point of intersection using elimination 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

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 \(`(4, 6)`\)

Shortest Distance From Point To a Line

• The shortest distance is always a straightline, thus, the shortest distance from a point to a line must be perpendicular.
• Thus, you can mind the slope of the line, then get the negative reciprocal (perpendicular slope), then find the equation of the perpendicular line.
• After you have the 2 lines, proceed by using subsitution or elimination to find the point of intersection.
• Then apply distance formula to find the shortest distance.