These eight questions follow geometry problems with several possible approaches. Write each answer on paper, with a one-sentence reason for your method, then open the answer.
Questions
1. Show that the triangle with P(−1, 2), Q(2, 6) and R(6, 3) has a right angle at Q.
Answer
Gradient of PQ = (6 − 2) ÷ (2 − (−1)) = 4/3. Gradient of QR = (3 − 6) ÷ (6 − 2) = −3/4. The product is −1, so PQ ⊥ QR.
Reason: coordinates are given as numbers and the question is about perpendicular lines, so gradients fit.
2. Choose coordinates or vectors, with a reason: (a) find the area of a quadrilateral from four vertices, (b) given OA = a and OB = b, show that M, the midpoint of AB, has position vector ½(a + b).
Answer
(a) Coordinates, because the area follows directly from the vertices. (b) Vectors, because the data is in letters with no coordinates.
3. A(1, 1), B(5, 2) and C(6, 6) are three vertices of parallelogram ABCD. Find D.
Answer
The diagonals of a parallelogram share a midpoint, so D = A + C − B = (1 + 6 − 5, 1 + 6 − 2) = (2, 5).
Check: the midpoint of AC is (3.5, 3.5), and the midpoint of BD is ((5 + 2) ÷ 2, (2 + 5) ÷ 2) = (3.5, 3.5).
4. Find the area of parallelogram ABCD from question 3.
Answer
Vertices in order: A(1, 1), B(5, 2), C(6, 6), D(2, 5). First sum: 1 × 2 + 5 × 6 + 6 × 5 + 2 × 1 = 64. Second sum: 1 × 5 + 2 × 6 + 6 × 2 + 5 × 1 = 34.
Area = ½ × |64 − 34| = 15 square units. Check: AB = (4, 1) and AD = (1, 4), and 4 × 4 − 1 × 1 = 15.
5. The segment AB has A(1, 2) and B(4, 8), on the line y = 2x. Which of the lines 2x + y = 10 and x + y = 30 meets the segment?
Answer
For 2x + y = 10: 2x + 2x = 10, so x = 2.5 and y = 5. Since 1 ≤ 2.5 ≤ 4, this line meets the segment at (2.5, 5).
For x + y = 30: 3x = 30, so x = 10 and y = 20. Since 10 is outside 1 ≤ x ≤ 4, this line does not meet the segment.
6. The line y = x − 2 meets the curve y = x² − 3x + 1. Find the point with positive y-coordinate.
Answer
Set equal: x² − 4x + 3 = 0, so (x − 1)(x − 3) = 0 and x = 1 or 3. The points are (1, −1) and (3, 1).
The point with positive y is (3, 1). Check: the curve gives 9 − 9 + 1 = 1 and the line gives 3 − 2 = 1.
7. A(−2, 3) and B(6, −1). The point P on AB has AP : PB = 3 : 1. Find P and the equation of the line through P perpendicular to AB.
Answer
P = ((1 × (−2) + 3 × 6) ÷ 4, (1 × 3 + 3 × (−1)) ÷ 4) = (4, 0). Vector check: A + ¾AB = (−2, 3) + ¾(8, −4) = (4, 0).
Gradient of AB = −4 ÷ 8 = −1/2, so the perpendicular gradient is 2. The line is y = 2(x − 4), so y = 2x − 8.
8. Write the one-sentence reason for using coordinates in question 7.
Answer
A model reason: “Coordinates are given as numbers and the question asks for the equation of a line, which needs a gradient and a point.” Any reason that links the wording to the method earns the mark.
If you got these wrong
- Questions 1 to 3 and 7: revisit choosing a coordinate method rather than a vector method.
- Questions 5, 6: read checking an intersection against a restriction.
- Question 4: recheck areas from coordinates.
Use the timed practice builder and a mistake log to keep practising. A teacher can go through your reasons in online one-to-one Additional Mathematics tuition.