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Mathematics β€” ALGEBRA

Subtopic: Equation of Straight Lines (Age ~14, Kenya)

In algebra, a straight line on the coordinate plane can be described by an equation. These notes explain common forms, how to find the equation from points or slope, and how to sketch a line.

Coordinate plane quick recap:
  • Horizontal axis = x-axis, vertical axis = y-axis.
  • Each point has coordinates written as (x, y).

1. Slope (Gradient) of a line

The slope m tells how steep a line is. For two points (x1, y1) and (x2, y2):

m = (y2 βˆ’ y1) / (x2 βˆ’ x1)

Example: points (1, 2) and (4, 5)

m = (5 βˆ’ 2) / (4 βˆ’ 1) = 3 / 3 = 1

2. Common forms of a straight line

a) Slope-intercept form
y = mx + c
- m is the slope (gradient).
- c is the y-intercept (point where the line crosses the y-axis).
Example: y = 2x + 1 β†’ slope m = 2, y-intercept c = 1 (point (0,1))
b) Point-slope form
y βˆ’ y1 = m(x βˆ’ x1)
- Use when you know a point (x1, y1) and the slope m.
Example: slope 3 through point (2, βˆ’1): y + 1 = 3(x βˆ’ 2).
c) Two-point form
If you know two points (x1, y1) and (x2, y2):
(y βˆ’ y1) / (x βˆ’ x1) = (y2 βˆ’ y1) / (x2 βˆ’ x1)
Rearranged it becomes point-slope with slope computed from the two points.
d) General form
ax + by + c = 0 (a, b, c are constants).
You can rearrange between general form and slope-intercept: y = mx + c.

3. Finding the equation from two points β€” worked example

Find the equation of the line through A(1, 2) and B(4, 5).

  1. Find slope: m = (5 βˆ’ 2)/(4 βˆ’ 1) = 3/3 = 1.
  2. Use point-slope with point A(1,2): y βˆ’ 2 = 1(x βˆ’ 1).
  3. Simplify: y βˆ’ 2 = x βˆ’ 1 β‡’ y = x + 1.

4. Intercepts

  • y-intercept: set x = 0, solve for y (gives point (0, c) in y = mx + c).
  • x-intercept: set y = 0, solve for x (solve 0 = mx + c β‡’ x = βˆ’c/m if m β‰  0).

5. Parallel and perpendicular lines

  • Parallel lines have the same slope. Example: y = 2x + 1 and y = 2x βˆ’ 3 are parallel.
  • Perpendicular lines have slopes that multiply to βˆ’1. If one slope is m, the perpendicular slope is βˆ’1/m. Example: slope 2 β†’ perpendicular slope βˆ’1/2.

6. Simple graph illustration

Below is a small sketch of the line y = 0.5x + 1 (for visual understanding):

βˆ’8 βˆ’4 0 4 8 15 10 5 0 βˆ’5 (0,1)

7. Short practice (try these)

  1. Find the slope of the line through (2, 3) and (5, 9).
  2. Write the equation of the line with slope βˆ’2 that passes through (1, 4).
  3. Find the equation of the line through (0, βˆ’2) and (3, 1).
  4. Are the lines y = 3x + 5 and y = 3x βˆ’ 2 parallel, perpendicular or neither?
Answers (check after trying):
  1. Slope = (9 βˆ’ 3)/(5 βˆ’ 2) = 6/3 = 2.
  2. Use y βˆ’ 4 = βˆ’2(x βˆ’ 1) β‡’ y βˆ’ 4 = βˆ’2x + 2 β‡’ y = βˆ’2x + 6.
  3. Slope = (1 βˆ’ (βˆ’2)) / (3 βˆ’ 0) = 3/3 = 1 β‡’ y = x βˆ’ 2.
  4. Parallel (same slope 3).

8. Tips for exams (KCSE/School tests)

  • Always calculate slope carefully; swap subtraction order consistently (y2βˆ’y1)/(x2βˆ’x1).
  • To graph quickly from y = mx + c: start at (0, c) then use rise/run = m.
  • Check special cases: vertical lines (x = k) have no slope (undefined); horizontal lines (y = k) have slope 0.

If you want, I can make more worked examples, printable exercise sheets, or short quizzes on equations of straight lines tailored to the Kenyan syllabusβ€”tell me which you prefer.

πŸ“ Practice Quiz

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