Coordinate Geometry

The line passes through the points A(1, 4) and B(5, 16). Find the gradient of the line.

• A3✓
• B12
• C4
• D1/3

Answer: Gradient = (16 − 4)/(5 − 1) = 12/4 = 3.

A circle has equation (x − 2)² + (y + 3)² = 25. State the coordinates of its centre.

• A(−2, −3)
• B(2, −3)✓
• C(−2, 3)
• D(2, 3)

Answer: For (x − a)² + (y − b)² = r², the centre is (a, b); here a = 2 and b = −3, so the centre is (2, −3).

Find the equation of the line through (0, −2) that is parallel to y = 3x + 7.

• Ay = 3x + 2
• By = −3x − 2
• Cy = (−1/3)x − 2
• Dy = 3x − 2✓

Answer: Parallel lines share the gradient 3, and the y-intercept is −2, giving y = 3x − 2.

A circle has equation x² + y² = 169. Find the length of its radius.

• A26
• B13✓
• C169
• D12

Answer: Comparing with x² + y² = r² gives r² = 169, so the radius r = √169 = 13.

A line has equation y = 2x − 5. Find the gradient of any line perpendicular to it.

• A2
• B−2
• C−1/2✓
• D1/2

Answer: Perpendicular gradients multiply to −1. The given gradient is 2, so the perpendicular gradient is −1/2.

Find the coordinates of the midpoint of the line segment joining A(−3, 7) and B(5, −1).

• A(1, 3)✓
• B(4, 4)
• C(2, 6)
• D(−4, 4)

Answer: The midpoint is ((−3 + 5)/2, (7 + (−1))/2) = (2/2, 6/2) = (1, 3).

Find the distance between the points P(2, 3) and Q(7, 15).

• A17
• B13✓
• C√17
• D12

Answer: Distance = √[(7 − 2)² + (15 − 3)²] = √(5² + 12²) = √(25 + 144) = √169 = 13.

A circle has equation x² + y² − 6x + 4y − 12 = 0. Find its radius by completing the square.

• A12
• B√12
• C25
• D5✓

Answer: Completing the square: (x − 3)² − 9 + (y + 2)² − 4 − 12 = 0, so (x − 3)² + (y + 2)² = 25. Hence r² = 25 and r = 5.

Find the equation of the line through the point (2, 7) that is perpendicular to y = (1/3)x + 1.

• Ay = −3x + 13✓
• By = (1/3)x + 7
• Cy = 3x + 1
• Dy = −3x + 7

Answer: The perpendicular gradient is −3 (negative reciprocal of 1/3). Using y − 7 = −3(x − 2) gives y = −3x + 6 + 7 = −3x + 13.

A circle has centre (1, 2). Find the perpendicular distance from the centre to the line 3x + 4y − 30 = 0.

• A19
• B19/25
• C3.8✓
• D5

Answer: Perpendicular distance = |3(1) + 4(2) − 30| / √(3² + 4²) = |−19| / 5 = 19/5 = 3.8.

The line y = x + k is a tangent to the curve y = x² + 3x + 4. Find the value of k.

• A4
• B3✓
• C−3
• D1

Answer: Setting x² + 3x + 4 = x + k gives x² + 2x + (4 − k) = 0. For a tangent the discriminant is 0: 2² − 4(4 − k) = 0, so 4 − 16 + 4k = 0, giving 4k = 12 and k = 3.

A curve is given parametrically by x = 2t and y = t² − 1. Find the Cartesian equation by eliminating the parameter t.

• Ay = 4x² − 1
• By = 2x² − 1
• Cy = x² − 1
• Dy = x²/4 − 1✓

Answer: From x = 2t we get t = x/2. Substituting into y = t² − 1 gives y = (x/2)² − 1 = x²/4 − 1.