These eight questions cover the whole coordinate geometry chapter, from the basic formulas to locus. All numbers are original, and each answer shows the working and a check.
Do them on paper first. Then use the mistake log and paper-error review to note which step went wrong on any question you missed.
Questions and answers
Question 1. Find the gradient, the midpoint and the length of the line segment joining A(−2, 3) and B(4, 11).
Answer
Gradient = (11 − 3) ÷ (4 + 2) = 8 ÷ 6 = 4/3.
Midpoint = ((−2 + 4) ÷ 2, (3 + 11) ÷ 2) = (1, 7).
Length = √(6² + 8²) = √100 = 10.
Question 2. Find the equation of the line through (2, 5) that is parallel to 3x + y = 4.
Answer
Rewrite as y = −3x + 4, so the gradient is −3. A parallel line has the same gradient.
y − 5 = −3(x − 2), so y = −3x + 11. Check: at x = 2, y = −6 + 11 = 5.
Question 3. Find the equation of the line through (4, −1) that is perpendicular to 2x − 3y = 6.
Answer
Rewrite as y = (2/3)x − 2, so the gradient is 2/3. The perpendicular gradient is −3/2.
y + 1 = −(3/2)(x − 4). Multiply by 2: 2y + 2 = −3x + 12, so 3x + 2y = 10. Check: 3(4) + 2(−1) = 10.
Question 4. Find the area of the triangle with vertices A(1, 1), B(7, 3) and C(4, 8).
Answer
Write the coordinates in a column and repeat the first: (1, 1), (7, 3), (4, 8), (1, 1).
Sum of downward products: 1 × 3 + 7 × 8 + 4 × 1 = 3 + 56 + 4 = 63.
Sum of upward products: 1 × 7 + 3 × 4 + 8 × 1 = 7 + 12 + 8 = 27.
Area = 1/2 × |63 − 27| = 1/2 × 36 = 18 square units.
Question 5. The points P(1, 2), Q(3, k) and R(7, 14) lie on a straight line. Find k.
Answer
Gradient of PR = (14 − 2) ÷ (7 − 1) = 12 ÷ 6 = 2.
Gradient of PQ must also be 2: (k − 2) ÷ (3 − 1) = 2, so k − 2 = 4 and k = 6.
Check with QR: (14 − 6) ÷ (7 − 3) = 2.
Question 6. A point P(x, y) moves so that it is always the same distance from A(0, 3) and B(4, −1). Find the equation of its locus.
Answer
PA² = PB², so x² + (y − 3)² = (x − 4)² + (y + 1)².
Expand: x² + y² − 6y + 9 = x² − 8x + 16 + y² + 2y + 1, so 8x − 8y − 8 = 0, which gives y = x − 1.
Check with the midpoint (2, 1): 1 = 2 − 1. The locus is the perpendicular bisector of AB.
Question 7. A point P moves so that its distance from C(2, 1) is always 5 units. Find the equation of its locus.
Answer
PC² = 25, so (x − 2)² + (y − 1)² = 25.
Expand: x² − 4x + 4 + y² − 2y + 1 = 25, so x² + y² − 4x − 2y − 20 = 0.
The locus is a circle with centre (2, 1) and radius 5.
Question 8. The point P divides the line joining A(−1, 4) and B(9, −1) in the ratio 2 : 3. Find the coordinates of P.
Answer
P = ((3 × (−1) + 2 × 9) ÷ 5, (3 × 4 + 2 × (−1)) ÷ 5) = (15 ÷ 5, 10 ÷ 5) = (3, 2).
Check: AB has components (10, −5). Two-fifths of that is (4, −2), and A + (4, −2) = (3, 2).
If you got these wrong
- Questions 1 to 3 need using gradient, midpoint and distance together and equations of parallel and perpendicular lines.
- Questions 4 and 5 need calculating areas from coordinates.
- Questions 6 and 7 need solving coordinate locus problems.
Ready for harder, mixed questions? Try geometry problems with several possible approaches. If you want a teacher to watch your working on these, see online one-to-one Additional Mathematics tuition.