How to Solve Coordinate Geometry Questions: ECZ Maths Tutorial
Coordinate Geometry is a compulsory topic in Examination Council of Zambia (ECZ) Mathematics across Grade 9 (Paper 1 & Paper 2) and Grade 12 / GCE (Paper 1 non-calculator & Paper 2 calculator).
Questions in this topic test your ability to work with Cartesian coordinates $(x, y)$, calculate gradients, find midpoints, determine line segment lengths, and construct equations of straight lines that are parallel or perpendicular to a given line. This tutorial breaks down every key formula with step-by-step worked ECZ past paper examples.
1. Gradient: $m = \frac{y_2 - y_1}{x_2 - x_1}$
2. Midpoint: $M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$
3. Distance: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
4. Line Equation: $y - y_1 = m(x - x_1)$
1. Worked Example: Gradient & Straight Line Equation
Worked Example 1
Question: A straight line passes through points $A(2, 5)$ and $B(6, 13)$. Find:
a) The gradient of line $AB$.
b) The equation of line $AB$ in the form $y = mx + c$.
Solution Step-by-Step:
- Let $(x_1, y_1) = (2, 5)$ and $(x_2, y_2) = (6, 13)$.
- Calculate gradient $m = \frac{13 - 5}{6 - 2} = \frac{8}{4} = \mathbf{2}$.
- Use point-slope formula $y - y_1 = m(x - x_1)$ with point $(2, 5)$:
$y - 5 = 2(x - 2) \implies y - 5 = 2x - 4 \implies \mathbf{y = 2x + 1}$.
2. Worked Example: Perpendicular Line Equation
Worked Example 2
Question: Find the equation of the line passing through point $P(3, -1)$ that is perpendicular to the line $y = 3x + 4$.
Solution Step-by-Step:
- The given line has gradient $m_1 = 3$.
- The perpendicular line gradient $m_2 = -\frac{1}{m_1} = \mathbf{-\frac{1}{3}}$.
- Substitute $m_2 = -\frac{1}{3}$ and point $P(3, -1)$ into $y - y_1 = m(x - x_1)$:
$y - (-1) = -\frac{1}{3}(x - 3) \implies y + 1 = -\frac{1}{3}x + 1 \implies \mathbf{y = -\frac{1}{3}x}$.
Frequently Asked Questions (FAQs)
How do you find the perpendicular gradient of a line in ECZ Mathematics?
If line 1 has a gradient m1, the perpendicular line 2 has a gradient m2 given by the negative reciprocal: m2 = -1 / m1 (since m1 * m2 = -1).
What is the formula for the distance between two points A(x1, y1) and B(x2, y2)?
The distance formula is d = sqrt((x2 - x1)^2 + (y2 - y1)^2).
What is the midpoint formula?
The midpoint M of a line segment connecting (x1, y1) and (x2, y2) is M = ((x1 + x2)/2, (y1 + y2)/2).