MYP9 Unit 5: Coordinate Geometry
1. Pythagorean Theorem
Definition
The Pythagorean Theorem states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. If \(a\) and \(b\) are the lengths of the legs of the triangle and \(c\) is the length of the hypotenuse, then the theorem can be written as:
\(a^2 + b^2 = c^2\)
Example
Problem: Given a right-angled triangle with legs of lengths 6 units and 8 units, find the length of the hypotenuse.
Solution:
- Identify the lengths of the legs: Let \(a = 6\) units and \(b = 8\) units.
- Apply the Pythagorean Theorem:
- Solve for \(c\):
\(a^2 + b^2 = c^2\)
\(6^2 + 8^2 = c^2\)
\(36 + 64 = c^2\)
\(100 = c^2\)
\(c = \sqrt{100}\)
\(c = 10\)
The length of the hypotenuse is 10 units.
2. Converse of the Pythagorean Theorem
Definition
The converse of the Pythagorean Theorem states that if the square of the length of one side of a triangle is equal to the sum of the squares of the lengths of the other two sides, then the triangle is a right-angled triangle. In other words, if \(a\), \(b\), and \(c\) are the lengths of the sides of a triangle and \(a^2 + b^2 = c^2\), then the triangle is right-angled.
Example
Problem: Determine whether a triangle with sides of lengths 5 units, 12 units, and 13 units is a right-angled triangle.
Solution:
- Identify the lengths of the sides: Let \(a = 5\) units, \(b = 12\) units, and \(c = 13\) units.
- Check the left-hand side (LHS) of the equation:
- Check the right-hand side (RHS) of the equation:
LHS = \(a^2 + b^2\)
LHS = \(5^2 + 12^2\)
LHS = \(25 + 144\)
LHS = \(169\)
RHS = \(c^2\)
RHS = \(13^2\)
RHS = \(169\)
Since LHS equals RHS, the triangle is a right-angled triangle.
3. Distance Between Two Points
Formula
If \((x_1, y_1)\) and \((x_2, y_2)\) are the coordinates of the two points, the distance \(d\) between these points is given by the formula:
\( d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2} \)
Example
Problem: Find the distance between the points \(A(3, 4)\) and \(B(7, 1)\).
Solution:
- Identify the coordinates of the points: Let \((x_1, y_1) = (3, 4)\) and \((x_2, y_2) = (7, 1)\).
- Apply the distance formula:
\( d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2} \)
\( d = \sqrt{(7 – 3)^2 + (1 – 4)^2} \)
\( d = \sqrt{4^2 + (-3)^2} \)
\( d = \sqrt{16 + 9} \)
\( d = \sqrt{25} \)
d = 5
The distance between the points \(A\) and \(B\) is 5 units.
4. Solve Systems of Equations Graphically
Formula
The midpoint formula is used to find the exact center point between two defined points in a coordinate plane. The midpoint of a line segment joining two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:
\( M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \)
Example
Problem: Find the midpoint of the line segment joining the points \(A(3, 6)\) and \(B(9, 18)\).
Solution:
- Identify the coordinates of the points:
- Apply the midpoint formula:
- Substitute the coordinates into the formula:
- Simplify the expression:
\( A(x_1, y_1) = (3, 6) \)
\( B(x_2, y_2) = (9, 18) \)
\( M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \)
\( M = \left( \frac{3 + 9}{2}, \frac{6 + 18}{2} \right) \)
\( M = \left( \frac{12}{2}, \frac{24}{2} \right) \)
\( M = (6, 12) \)
The midpoint of the line segment joining the points \(A(3, 6)\) and \(B(9, 18)\) is \((6, 12)\).
5. The Gradient (or Slope)
Formula
The gradient (or slope) of a line is a measure of how steep the line is. It is calculated as the ratio of the vertical change to the horizontal change between two points on the line. The formula for the gradient between two points \((x_1, y_1)\) and \((x_2, y_2)\) is:
\( m = \frac{y_2 – y_1}{x_2 – x_1} \)
Types of Slope
| Description | Slope |
|---|---|
| A line sloping upwards from left to right has a positive gradient. | \(m\) is positive |
| A line sloping downwards from left to right has a negative gradient. | \(m\) is negative |
| Parallel lines have the same gradient. | \(m_1 = m_2\) |
| A horizontal line has a gradient of 0. | \(m = 0\) |
| The gradient of a vertical line is undefined. | \(m\) is undefined |
Example
Problem: Find the slope of the line passing through the points \(P(2, 3)\) and \(Q(5, 11)\).
Solution:
- Identify the coordinates of the points:
- Apply the slope formula:
- Substitute the coordinates into the formula:
- Simplify the expression:
\( P(x_1, y_1) = (2, 3) \)
\( Q(x_2, y_2) = (5, 11) \)
\( m = \frac{y_2 – y_1}{x_2 – x_1} \)
\( m = \frac{11 – 3}{5 – 2} \)
\( m = \frac{8}{3} \)
The slope of the line passing through the points \(P(2, 3)\) and \(Q(5, 11)\) is \(\frac{8}{3}\).