Work through the levels in order. Type your answer and press Check — open the worked solution if you get stuck. Answers like -6, 3/4 or 0.25 are typed exactly as shown. A ⭐ Bonus Challenge at the end is extra credit — skip it if the rest was hard work.
Level 1 · Warm-up
1 Find the gradient of the line through U(−3, −2) and V(5, 4).
Show worked solution
Rise = 4 − (−2) = 6 and run = 5 − (−3) = 8, so m = 6 ÷ 8 = \(\dfrac{3}{4}\). \(\dfrac{4}{3}\) is run over rise (inverted); −\(\dfrac{3}{4}\) has the wrong sign, though the line rises to the right; −\(\dfrac{4}{3}\) makes both errors.
2 Find the midpoint of the interval joining (−6, 11) and (4, −3).
Show worked solution
x: (−6 + 4) ÷ 2 = −1. y: (11 + (−3)) ÷ 2 = 4. The midpoint is (−1, 4).
3 Find the length of the interval from (−3, −4) to (4, 20).
Show worked solution
Run 7 and rise 24. 72 + 242 = 49 + 576 = 625, and √625 = 25.
Level 2 · Core
4 A is (−1, −2) and B is (3, 1). Which point lies on the line AB?
Show worked solution
m(AB) = 3 ÷ 4: across 4, up 3. From B(3, 1), across 4 and up 3 reaches (7, 4). (6, 5) is across 3 and up 4 from B, the gradient upside down (\(\dfrac{4}{3}\)); (−5, 1) is across −4 and up 3 from A, gradient −\(\dfrac{3}{4}\), a sign error; (4, 2) is across 1 and up 1 from B, gradient 1.
5 M(2, −3) is the midpoint of PQ, and P is (−5, 1). Find Q.
Show worked solution
P to M: x goes −5 → 2 (+7) and y goes 1 → −3 (−4). Again from M: x = 2 + 7 = 9 and y = −3 − 4 = −7, so Q is (9, −7). Check: ((−5 + 9) ÷ 2, (1 + (−7)) ÷ 2) = (2, −3) ✓.
6 The line through A(2, k) and B(6, 1) has gradient −\(\dfrac{3}{2}\). Find k.
Show worked solution
(1 − k) ÷ 4 = −3/2, so 1 − k = −6 and k = 7. Check: (1 − 7) ÷ (6 − 2) = −6 ÷ 4 = −\(\dfrac{3}{2}\) ✓.
Level 3 · Challenge
7 Three statements:
1 The interval from (−4, −1) to (2, −1) has a gradient that is not defined.
2 The points (0, −4), (3, 2) and (5, 6) lie on one straight line.
3 The interval from (−2, 7) to (6, −8) is longer than 16 units.
Which statements are correct?
1 The interval from (−4, −1) to (2, −1) has a gradient that is not defined.
2 The points (0, −4), (3, 2) and (5, 6) lie on one straight line.
3 The interval from (−2, 7) to (6, −8) is longer than 16 units.
Which statements are correct?
Show worked solution
Both points in statement 1 have y = −1, so the interval is horizontal and its gradient is 0, a perfectly good number (1 ✗). The gradients 6 ÷ 3 = 2 and 4 ÷ 2 = 2 are equal, so the points are collinear (2 ✓). Run 8, rise −15: 64 + 225 = 289, so the length is 17, which is more than 16 (3 ✓). 2 and 3.
8 P(a, 5) and Q(10, b) have midpoint M(2, −1). Find a and b, and then the length of PQ.
Show worked solution
a + 10 = 4, so a = −6. 5 + b = −2, so b = −7. From P(−6, 5) to Q(10, −7): run 16, rise −12, and 256 + 144 = 400, so PQ = 20.
9 A is (−3, 4) and M(1, −3.5) is the midpoint of AB. How long is AB?
Show worked solution
A to M is +4 in x and −7.5 in y, so B = (1 + 4, −3.5 − 7.5) = (5, −11). From A to B: run 8, rise −15, so AB2 = 64 + 225 = 289 and AB = 17. Or: AM2 = 16 + 56.25 = 72.25, AM = 8.5, and AB = 2 × 8.5 = 17.