Coordinate Geometry Notes - Grade 11 Mathematics | YNetStudyHub
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Mathematics Tutorial

Coordinate Geometry

Lesson 81 of 87
3 min read Mathew Wahome

Introduction

Coordinate geometry is a branch of mathematics that deals with the study of geometric figures using coordinates. It involves representing points on a plane using ordered pairs of numbers, usually denoted as $(x, y)$, where $x$ represents the horizontal position (abscissa) and $y$ represents the vertical position (ordinate). By using coordinates, we can determine properties of geometric shapes, such as distances between points, slopes of lines, and equations of curves.

Distance Formula

The distance between two points $A(x_1, y_1)$ and $B(x_2, y_2)$ in a coordinate plane can be calculated using the distance formula: $$ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$

Example: Find the distance between points $A(3, 5)$ and $B(7, 9)$. $$ \text{Distance} = \sqrt{(7 - 3)^2 + (9 - 5)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} $$

Slope of a Line

The slope of a line passing through two points $A(x_1, y_1)$ and $B(x_2, y_2)$ is given by: $$ \text{Slope} = \frac{y_2 - y_1}{x_2 - x_1} $$

Example: Find the slope of the line passing through points $A(2, 4)$ and $B(6, 10)$. $$ \text{Slope} = \frac{10 - 4}{6 - 2} = \frac{6}{4} = \frac{3}{2} $$

Midpoint Formula

The coordinates of the midpoint of a line segment with endpoints $A(x_1, y_1)$ and $B(x_2, y_2)$ can be found using the midpoint formula: $$ \text{Midpoint} = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) $$

Example: Find the midpoint of the line segment with endpoints $A(-1, 3)$ and $B(5, -1)$. $$ \text{Midpoint} = \left(\frac{-1 + 5}{2}, \frac{3 + (-1)}{2}\right) = (2, 1) $$

Equation of a Line

The equation of a line in the form $y = mx + c$, where $m$ is the slope and $c$ is the y-intercept. If we know a point $(x_1, y_1)$ on the line, we can find the equation using: $$ y - y_1 = m(x - x_1) $$

Example: Find the equation of the line passing through point $(-2, 4)$ with slope $\frac{1}{3}$. $$ y - 4 = \frac{1}{3}(x + 2) \Rightarrow y = \frac{1}{3}x + \frac{10}{3} $$

Circle Equation

The equation of a circle with center $(h, k)$ and radius $r$ is given by: $$ (x - h)^2 + (y - k)^2 = r^2 $$

Example: Find the equation of a circle with center at $(2, -3)$ and radius 5. $$ (x - 2)^2 + (y + 3)^2 = 25 $$

Common Mistakes

  • Forgetting to square the differences in the distance formula.
  • Incorrectly calculating the slope by swapping the $x$ and $y$ coordinates.
  • Misinterpreting the midpoint formula and not dividing by 2.

Key Points

  • Coordinate geometry uses ordered pairs of numbers to represent points on a plane.
  • The distance formula calculates the distance between two points.
  • The slope formula determines the inclination of a line.
  • The midpoint formula finds the middle point of a line segment.
  • Equations of lines and circles can be derived using specific formulas.

Practice Questions

  1. Find the distance between points $P(1, -2)$ and $Q(4, 5)$.
  2. Determine the slope of the line passing through points $C(-3, 2)$ and $D(1, 8)$.
  3. Calculate the midpoint of the line segment with endpoints $E(0, -4)$ and $F(6, 2)$.
  4. Write the equation of the line with slope $\frac{2}{3}$ passing through point $G(4, -1)$.
  5. Find the equation of a circle with center at $(3, -1)$ and radius 4.

Practice Questions - Answers

  1. The distance between $P(1, -2)$ and $Q(4, 5)$ is: $$ \sqrt{(4 - 1)^2 + (5 - (-2))^2} = \sqrt{9 + 49} = \sqrt{58} $$
  2. The slope of the line passing through $C(-3, 2)$ and $D(1, 8)$ is: $$ \frac{8 - 2}{1 - (-3)} = \frac{6}{4} = \frac{3}{2} $$
  3. The midpoint of $E(0, -4)$ and $F(6, 2)$ is: $$ \left(\frac{0 + 6}{2}, \frac{-4 + 2}{2}\right) = (3, -1) $$
  4. The equation of the line passing through $G(4, -1)$ with slope $\frac{2}{3}$ is: $$ y - (-1) = \frac{2}{3}(x - 4) \Rightarrow y = \frac{2}{3}x + \frac{5}{3} $$
  5. The equation of the circle with center at $(3, -1)$ and radius 4 is: $$ (x - 3)^2 + (y + 1)^2 = 16 $$
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