Last reviewed: 7 November 2026
Updated November 2026. These Grade 9 equation of a straight line notes show how to find the gradient of a line, write its equation in the form y = mx + c, and read the gradient and y-intercept straight from an equation. In short: the gradient m is the change in y divided by the change in x between two points, and c is where the line cuts the y-axis. Once you know m and one point, you can always write the equation.
The examples and questions are original practice for Kenyan junior school learners, not an official paper. For the other Grade 9 maths topics, use our Grade 9 maths revision guide and the Grade 9 notes hub. If solving simple equations is still shaky, revise linear equations for junior school first.
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Where This Topic Sits in Grade 9 Maths
The Grade 9 Mathematics curriculum design from the Kenya Institute of Curriculum Development lists “Equation of a Straight Line” as a sub-strand under Algebra, with the largest lesson allocation in that strand. It covers gradient, the equation of a line in different forms, the x- and y-intercepts, and the gradient-intercept form y = mx + c. The designs are listed on the KICD Grade Nine curriculum designs page. Parallel and perpendicular line conditions are left out here because they belong to later work.
What Is the Gradient of a Line?
Gradient measures steepness. For two points (x1, y1) and (x2, y2):
gradient m = (y2 − y1) ÷ (x2 − x1)
This is the “rise over run” idea explained in OpenStax Prealgebra’s section on the slope of a line (slope is another name for gradient).
| Line goes… | Gradient is… | Example |
|---|---|---|
| Up from left to right | Positive | Through (0, 1) and (2, 5): m = 4 ÷ 2 = 2 |
| Down from left to right | Negative | Through (0, 6) and (3, 0): m = −6 ÷ 3 = −2 |
| Flat (horizontal) | Zero | Through (1, 4) and (5, 4): m = 0 ÷ 4 = 0 |
| Straight up (vertical) | Undefined | Through (3, 1) and (3, 7): division by zero |
Worked example: gradient from two points
Find the gradient of the line through A(−1, 2) and B(3, 10).
- Change in y: 10 − 2 = 8.
- Change in x: 3 − (−1) = 4.
- Gradient: 8 ÷ 4 = 2.
Keep the order the same on top and bottom. If you start with B’s y-value, start with B’s x-value too. Mixing the order flips the sign and is the most common error in this topic.
Writing the Equation of a Straight Line
Method 1: Gradient and one point
Take a general point (x, y) on the line. Its gradient with the known point must equal m. Example: a line has gradient 3 and passes through (2, 5).
- (y − 5) ÷ (x − 2) = 3
- y − 5 = 3(x − 2) = 3x − 6
- y = 3x − 1
Check: put x = 2 into y = 3x − 1. You get y = 5, which matches the given point.
Method 2: Two points
Find the gradient first, then use Method 1. For A(−1, 2) and B(3, 10) the gradient is 2. Using A: y − 2 = 2(x + 1), so y = 2x + 4. Check with B: 2(3) + 4 = 10. Correct.
The Form y = mx + c
When the equation is written as y = mx + c, the number multiplying x is the gradient and c is the y-intercept, the value of y where the line crosses the y-axis (where x = 0).
- y = 4x − 3: gradient 4, y-intercept −3.
- y = −x + 7: gradient −1, y-intercept 7.
- y = 5: gradient 0, y-intercept 5 (a horizontal line).
If the equation is not in this form, rearrange it first. Example: 2y − 6x = 8. Add 6x to both sides: 2y = 6x + 8. Divide by 2: y = 3x + 4. The gradient is 3 and the y-intercept is 4. Reading “−6” as the gradient before rearranging is a frequent mistake.
Finding the x- and y-Intercepts
To find where a line crosses the axes, set one variable to zero.
- y-intercept: put x = 0.
- x-intercept: put y = 0.
For 3x + 2y = 12: when x = 0, 2y = 12 so y = 6, giving (0, 6). When y = 0, 3x = 12 so x = 4, giving (4, 0). Plotting these two points and joining them is a quick way to draw the line on squared paper.
A Real-Life Example
A boda boda rider charges a fixed KSh 50 plus KSh 30 per kilometre in this made-up example. The cost y for x kilometres is y = 30x + 50. The gradient (30) is the cost per extra kilometre and the y-intercept (50) is the fixed charge before the trip starts. A 4 km trip costs 30 × 4 + 50 = KSh 170. Seeing gradient as a “rate” helps in science and business questions too.
Original Practice Questions With Worked Answers
- Find the gradient through (2, −3) and (6, 5). (5 − (−3)) ÷ (6 − 2) = 8 ÷ 4 = 2.
- Write the equation of the line with gradient −2 through (1, 4). y − 4 = −2(x − 1), so y = −2x + 6.
- State the gradient and y-intercept of 4y = 8x − 12. Divide by 4: y = 2x − 3. Gradient 2, y-intercept −3.
- Find the intercepts of 5x − y = 10. x = 0 gives y = −10, so (0, −10). y = 0 gives x = 2, so (2, 0).
- Does the point (3, 7) lie on y = 2x + 1? 2(3) + 1 = 7, so yes.
Try each question before reading the answer. For timed practice, use the Grade 9 revision papers or the Grade 9 CBC Challenge. These notes are free; full revision access is available with the KSh 20 Day Pass or the KSh 500 Monthly Pass.
Frequently Asked Questions
What is the formula for the gradient of a straight line?
Gradient = (y2 − y1) ÷ (x2 − x1), the change in y divided by the change in x between any two points on the line.
What do m and c mean in y = mx + c?
m is the gradient (steepness) of the line and c is the y-intercept, the point where the line crosses the y-axis.
How do I find the equation of a line from two points?
Find the gradient from the two points, then substitute the gradient and one of the points into y − y1 = m(x − x1) and simplify to y = mx + c.
What is the gradient of a horizontal line?
Zero. The y-value does not change, so the change in y is 0 and the gradient is 0.