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O-Level Geometry using coordinates
What the O-Level syllabus expects for Geometry using coordinates, and how to practise it.
What the syllabus expects
- obtaining the gradient of a straight line from the coordinates of two points lying on it
- computing how long a line segment is from the coordinates of the two ends
- reading off, and determining, the equation y = mx + c belonging to a straight-line graph
- solving geometry problems with the help of coordinates
How it's examined
Questions on this topic most often ask you to find, solve, determine.
Worked examples
Example 1 (2 marks)
Three collinear points are given as (3, 4), (2, k) and (7, 5). Work out the value of k.
Show the worked answer
Points (3, 4), (2, k) and (7, 5) are collinear, so gradients are equal. Gradient from (3,4) to (7,5) = (5 - 4)/(7 - 3) = 1/4. Gradient from (3,4) to (2,k) = (k - 4)/(2 - 3) = (k - 4)/(-1). Set equal: (k - 4)/(-1) = 1/4 k - 4 = -1/4 k = 4 - 1/4 = 3.75 = 15/4.
Example 2 (3 marks)
Find the equation of the line joining P(−6, 7) and R(−1, 1).
Show the worked answer
Step 1: Find the gradient of the line through the two points. The gradient formula is m = (y₂ − y₁)/(x₂ − x₁). Take P(−6, 7) as point 1 and R(−1, 1) as point 2: m = (1 − 7)/(−1 − (−6)) Careful with the double negative on the bottom: −1 − (−6) = −1 + 6 = 5 m = (−6)/5 = −6/5 Step 2: Use y = mx + c and one of the points to find the intercept c. Substitute P(−6, 7) and m = −6/5: 7 = (−6/5)(−6) + c (−6/5)(−6) = 36/5 7 = 36/5 + c c = 7 − 36/5 Write 7 as 35/5 so the fractions can be subtracted: c = 35/5 − 36/5 = −1/5 Step 3: Write down the equation. y = (−6/5)x − 1/5 Step 4: Check with the other point R(−1, 1). (−6/5)(−1) − 1/5 = 6/5 − 1/5 = 5/5 = 1 ✓, which is the y-coordinate of R. The equation of PR is y = −6/5 x − 1/5 (equivalently 6x + 5y + 1 = 0).
Example 3 (3 marks)
Two points, B(1, 0) and A(k, 2), are a distance of √(2k+2) units apart. Determine the possible values of k.
Show the worked answer
Step 1: Write the distance between the two points and square both sides. The distance between B(1, 0) and A(k, 2) is AB = √((k − 1)² + (2 − 0)²) We are told this distance equals √(2k + 2). Squaring both sides removes both square roots at once: (k − 1)² + (2 − 0)² = 2k + 2 Step 2: Expand the bracket and collect everything on one side. (k − 1)² = k² − 2k + 1, and (2 − 0)² = 4, so k² − 2k + 1 + 4 = 2k + 2 k² − 2k + 5 = 2k + 2 Subtract 2k and 2 from both sides: k² − 2k − 2k + 5 − 2 = 0 k² − 4k + 3 = 0 Step 3: Factorise the quadratic. We need two numbers that multiply to +3 and add to −4. Those numbers are −1 and −3. k² − 4k + 3 = (k − 1)(k − 3) = 0 A product is zero only when a factor is zero, so k − 1 = 0 giving k = 1, or k − 3 = 0 giving k = 3 Step 4: Check both values are usable, since √(2k + 2) must be a real, non-negative length. k = 1: 2k + 2 = 4, so the stated distance is √4 = 2; and (1 − 1)² + 2² = 0 + 4 = 4 ✓ k = 3: 2k + 2 = 8, so the stated distance is √8; and (3 − 1)² + 2² = 4 + 4 = 8 ✓ Both values work. k = 1 or k = 3.
More worked questions on this topic
- Points A(20, 10) and B(14, 0) lie in the plane. (i) Find the equation of line AB. (ii) If W(9, (4 marks)
More O-Level Elementary Maths (E-Maths) topics
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