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 What is the gradient of any line perpendicular to 10x + 6y = 1?
Show worked solution
6y = −10x + 1, so the gradient is −10 ÷ 6 = −\(\dfrac{5}{3}\). Flip it and change the sign: \(\dfrac{3}{5}\). Check: −\(\dfrac{5}{3}\) × \(\dfrac{3}{5}\) = −1 ✓. −\(\dfrac{5}{3}\) is the parallel gradient; −\(\dfrac{3}{5}\) flips but keeps the sign; \(\dfrac{5}{3}\) changes the sign but does not flip.
2 Find the line through (−4, 5) that is parallel to 6x + 3y = 2. Write it as y = mx + c.
Show worked solution
3y = −6x + 2, so the gradient is −2. Through (−4, 5): 5 = −2(−4) + c = 8 + c, so c = −3, and the line is y = −2x − 3.
3 The line through (2, k) and (5, 7) is perpendicular to a line with gradient \(\dfrac{3}{2}\). Find k.
Show worked solution
The perpendicular gradient is −\(\dfrac{2}{3}\). (7 − k) ÷ 3 = −\(\dfrac{2}{3}\), so 7 − k = −2 and k = 9. Check: (7 − 9) ÷ 3 = −\(\dfrac{2}{3}\) ✓.
Level 2 · Core
4 Find the line through Q(−3, −2) that is perpendicular to the line drawn through (0, 2) and (4, −1).
Show worked solution
The drawn line has gradient −3 ÷ 4 = −\(\dfrac{3}{4}\), so the perpendicular gradient is \(\dfrac{4}{3}\). y + 2 = \(\dfrac{4}{3}\)(x + 3), so 3y + 6 = 4x + 12, which is 4x − 3y = −6. Check Q: −12 + 6 = −6 ✓. 3x + 4y = −17 is parallel to the drawn line; 4x + 3y = −18 (gradient −\(\dfrac{4}{3}\)) flips but keeps the sign; 3x − 4y = −1 (gradient \(\dfrac{3}{4}\)) changes the sign but does not flip.
5 Line k is perpendicular to y = 3x − 4 and passes through (6, 1). Find where k crosses the y-axis and the x-axis.
Show worked solution
Gradient of k: −\(\dfrac{1}{3}\). y − 1 = −\(\dfrac{1}{3}\)(x − 6) = −\(\dfrac{1}{3}\)x + 2, so y = −\(\dfrac{1}{3}\)x + 3. The y-intercept is 3. Put y = 0: \(\dfrac{1}{3}\)x = 3, so the x-intercept is 9.
6 The lines 2x + ay = 5 and 3x − 4y = 1 are perpendicular. Find a.
Show worked solution
3x − 4y = 1 gives y = \(\dfrac{3}{4}\)x − \(\dfrac{1}{4}\), gradient \(\dfrac{3}{4}\), so the other line needs gradient −\(\dfrac{4}{3}\). 2x + ay = 5 gives y = −\(\dfrac{2}{a}\)x + \(\dfrac{5}{a}\). −\(\dfrac{2}{a}\) = −\(\dfrac{4}{3}\), so 4a = 6 and a = 1.5.
Level 3 · Challenge
7 Three statements:
1 A(−1, −1), B(2, 3) and C(−2, 6) make a right angle at B.
2 The lines y = \(\dfrac{2}{3}\)x + 1 and 2x + 3y = 6 are perpendicular.
3 The line through (−2, 3) parallel to the x-axis is y = 3.
Which statements are correct?
1 A(−1, −1), B(2, 3) and C(−2, 6) make a right angle at B.
2 The lines y = \(\dfrac{2}{3}\)x + 1 and 2x + 3y = 6 are perpendicular.
3 The line through (−2, 3) parallel to the x-axis is y = 3.
Which statements are correct?
Show worked solution
AB has gradient 4 ÷ 3 = \(\dfrac{4}{3}\) and BC has gradient 3 ÷ (−4) = −\(\dfrac{3}{4}\); their product is −1 (1 ✓). 2x + 3y = 6 has gradient −\(\dfrac{2}{3}\), and \(\dfrac{2}{3}\) × (−\(\dfrac{2}{3}\)) = −\(\dfrac{4}{9}\), not −1: that is the negative without the reciprocal (2 ✗). A line parallel to the x-axis is horizontal, and every point on it has y = 3 (3 ✓). 1 and 3.
8 ABCD is a rectangle with A(−2, −1), B(4, 1) and C(3, 4). Find D.
Show worked solution
B to C is −1 in x and +3 in y. The same step from A: D = (−2 − 1, −1 + 3) = (−3, 2). Check: AB has gradient 2 ÷ 6 = \(\dfrac{1}{3}\) and AD has gradient 3 ÷ (−1) = −3, and \(\dfrac{1}{3}\) × (−3) = −1, so the angle at A is a right angle ✓.
9 Triangle PQR has P(−1, 2), Q(4, 4) and R(k, −8). The angle at P is a right angle. Find k.
Show worked solution
Gradient PQ = (4 − 2) ÷ (4 − (−1)) = \(\dfrac{2}{5}\), so PR needs gradient −\(\dfrac{5}{2}\). (−8 − 2) ÷ (k + 1) = −\(\dfrac{5}{2}\), so −10 × 2 = −5(k + 1), k + 1 = 4 and k = 3. Check: PR goes from (−1, 2) to (3, −8): −10 ÷ 4 = −\(\dfrac{5}{2}\) ✓.