Mathematics — Geometry

Subtopic: Coordinates and Graphs (Age ≈ 14, Kenyan context)

Objectives: By the end of these notes you should be able to:

• Use and read coordinates on the Cartesian plane.
• Plot points and simple shapes from coordinates.
• Find distance and midpoint between two points.
• Find and use the gradient (slope) and equation of a straight line.

1. The Cartesian coordinate plane

- The plane has two perpendicular number lines: the x-axis (horizontal) and the y-axis (vertical).
- The point where they meet is the origin O(0, 0).
- A coordinate is written (x, y): x is distance from the y-axis (left negative, right positive), y is distance from the x-axis (down negative, up positive).

2. How to write and plot coordinates

1. Start at the origin (0,0).
2. Move along x (left for negative, right for positive) the value of x.
3. From there move along y (down for negative, up for positive) the value of y.
4. Mark the point and label it (x, y).

3. Distance between two points

If P(x1, y1) and Q(x2, y2), distance PQ is found using the distance formula:

PQ = sqrt((x2 - x1)^2 + (y2 - y1)^2)

Example: Find distance between A(2, 3) and B(-3, -2).

PQ = sqrt((−3 − 2)^2 + (−2 − 3)^2) = sqrt((−5)^2 + (−5)^2) = sqrt(25 + 25) = sqrt(50) = 5√2.

4. Midpoint formula

Midpoint M of PQ: M = ((x1 + x2)/2 , (y1 + y2)/2).

Example: Midpoint of A(2,3) and B(-3,-2):

M = ((2 + (−3))/2, (3 + (−2))/2) = (−1/2, 1/2).

5. Gradient (slope) of a line

Gradient m between P(x1,y1) and Q(x2,y2) is:

m = (y2 − y1) / (x2 − x1)

If m is positive the line rises to the right; if m is negative the line falls to the right. Vertical lines have undefined gradient; horizontal lines have m = 0.

Example: Gradient of AB where A(2,3) and B(-3,-2):

m = (−2 − 3)/(−3 − 2) = (−5)/(−5) = 1

6. Equation of a straight line

Common form: y = mx + c, where m = gradient and c = y-intercept (value of y when x = 0).

1. To find equation from m and one point (x1,y1): y − y1 = m(x − x1) (point-slope form). Rearrange to y = mx + c.
2. To graph y = mx + c: mark c on y-axis and use gradient m to find another point (rise/run).

Example: Find equation of line through A(2,3) and B(-3,-2).

We found m = 1. Use point A(2,3): y − 3 = 1(x − 2) → y − 3 = x − 2 → y = x + 1. So c = 1.

7. Special lines

• Vertical line: x = a (no y term). Example: x = 2 is vertical through x = 2 (undefined gradient).
• Horizontal line: y = b (gradient 0). Example: y = −1 is horizontal through y = −1.

8. How to graph from equation (easy steps)

1. If y = mx + c, plot point (0, c) on y-axis.
2. From (0, c) move right by 1 and up by m (if m positive), or down if m negative — repeat to get several points.
3. Join points with a straight line and extend both ways with arrowheads.
4. Label the equation on the graph.

9. Worked example: graph y = 2x − 1

- y-intercept c = −1 → plot (0, −1).
- Gradient m = 2 → rise 2, run 1. From (0, −1) go to (1, 1) and then (2, 3).
- Join the points and draw a straight line.

10. Practice questions

1. Plot points P(1,2), Q(4,2) and R(4,−1). Draw triangle PQR and calculate its area (use grid square method).
2. Find the distance and midpoint between C(−1, 4) and D(3, −2).
3. Find the gradient of the line joining E(0, −1) and F(5, 4).
4. Find the equation of the line through G(2, 0) with gradient m = −1/2.
5. Graph y = −x + 2 and state its y-intercept and gradient.
6. State the equation of the horizontal line through (−3, 5) and the vertical line through (2, −1).
7. Two points are (2, k) and (−4, 3). Find k if the gradient between them is −1.
8. Show that the triangle with vertices (0,0), (4,0) and (4,3) is a right triangle and find its area.

11. Answers & hints

1. Triangle P(1,2), Q(4,2), R(4,−1): Base PQ = 3, height PR = 3 → area = 1/2 × 3 × 3 = 4.5 square units.
2. Distance CD = sqrt((3−(−1))^2 + (−2−4)^2) = sqrt(16 + 36) = sqrt(52) = 2√13. Midpoint = ((−1+3)/2, (4+(−2))/2) = (1, 1).
3. m = (4 − (−1))/(5 − 0) = 5/5 = 1.
4. Use y − 0 = (−1/2)(x − 2) → y = −1/2 x + 1.
5. y = −x + 2: gradient m = −1, y-intercept = 2.
6. Horizontal: y = 5. Vertical: x = 2.
7. Gradient m = (3 − k)/(−4 − 2) = (3 − k)/(−6) = −1 → 3 − k = 6 → k = −3.
8. Vectors: (4,0) − (0,0) = (4,0), (4,3) − (0,0) = (4,3). Dot product = 4×4 + 0×3 = 16; but to test right angle check slopes: slope of base = 0, slope of vertical side = ∞ → one side horizontal, one vertical → right angle at (4,0). Area = 1/2 × base 4 × height 3 = 6.

12. Exam tips (Kenyan context)

• Always label axes and use a clear scale (1 grid = 1 unit is easiest).
• Write coordinates as (x, y) — x first, then y.
• When asked for gradient between two points, subtract in the same order: (y2 − y1)/(x2 − x1).
• Check special cases: if x1 = x2 gradient is undefined (vertical line), if y1 = y2 gradient = 0 (horizontal).
• Show working: examiners award marks for correct method even if arithmetic slips occur.

If you want, I can make printable A4 notes, extra practice questions, or step-by-step solutions for any of the practice questions above.