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.

Find the distance between the points A(1, 2) and B(4, 6).

• A5✓
• B7
• C√7
• D25

Answer: Distance = √((4 − 1)² + (6 − 2)²) = √(9 + 16) = √25 = 5. Adding the differences gives 3 + 4 = 7, and stopping before the square root leaves 25.

A circle has equation (x − 3)² + (y + 2)² = 25. State its centre and radius.

• ACentre (3, −2), radius 25
• BCentre (−3, 2), radius 5
• CCentre (3, −2), radius 5✓
• DCentre (−3, 2), radius 25

Answer: In (x − a)² + (y − b)² = r² the centre is (a, b) and the radius is r. Here a = 3 and b = −2, and r² = 25 so r = 5. The two traps are flipping the signs of the centre and quoting r² as the radius.

A circle has equation (x + 1)² + (y − 4)² = 9. State its centre and radius.

• ACentre (−1, 4), radius 3✓
• BCentre (1, −4), radius 3
• CCentre (−1, 4), radius 9
• DCentre (1, −4), radius 9

Answer: In (x − a)² + (y − b)² = r² the centre is (a, b) and the radius is r. Here x + 1 is x − (−1), so a = −1 and b = 4, and r² = 9 so r = 3. The two traps are flipping the signs of the centre and quoting r² instead of r.

A circle has equation (x − 2)² + (y − 6)² = 49. State its centre and radius.

• ACentre (2, 6), radius 7✓
• BCentre (−2, −6), radius 7
• CCentre (2, 6), radius 49
• DCentre (−2, −6), radius 49

Answer: In (x − a)² + (y − b)² = r² the centre is (a, b) and the radius is r. Here a = 2 and b = 6, and r² = 49 so r = 7. The two traps are flipping the signs of the centre and quoting r² as the radius.

Find the gradient of the line perpendicular to the line through (1, 2) and (5, 10).

• A2
• B−1/2✓
• C−2
• D1/2

Answer: The gradient of the line is (10 − 2)/(5 − 1) = 2. Perpendicular gradients multiply to −1, so the answer is −1/(2) = −1/2. Simply negating the gradient, or only taking the reciprocal, are the two usual errors.

Find the gradient of the line perpendicular to the line through (−2, 3) and (4, −9).

• A−2
• B1/2✓
• C2
• D−1/2

Answer: The gradient of the line is (−9 − 3)/(4 − (−2)) = −2. Perpendicular gradients multiply to −1, so the answer is −1/(−2) = 1/2. Simply negating the gradient, or only taking the reciprocal, are the two usual errors.

Find the gradient of the line perpendicular to the line through (0, −1) and (6, 11).

• A−1/2✓
• B2
• C−2
• D1/2

Answer: The gradient of the line is (11 − (−1))/(6 − 0) = 2. Perpendicular gradients multiply to −1, so the answer is −1/(2) = −1/2. Simply negating the gradient, or only taking the reciprocal, are the two usual errors.

Find the distance between (1, 2) and (7, 10).

• A6
• B14
• C100
• D10✓

Answer: Distance = √[(7 − 1)² + (10 − 2)²] = √100 = 10. The value before square-rooting is the usual wrong answer.

Find the distance between (−3, 4) and (5, −2).

• A10✓
• B14
• C100
• D8

Answer: Distance = √[(5 − −3)² + (−2 − 4)²] = √100 = 10. The value before square-rooting is the usual wrong answer.

The line y = 2x + 1 meets the circle x² + y² = 25. How is the number of intersection points found?

• AFind the centre and check it lies on the line
• BCompare the gradient of the line with the radius
• CSubstitute for y and use the discriminant of the resulting quadratic✓
• DDifferentiate the circle equation and set it equal to 2

Answer: Substituting y = 2x + 1 into the circle gives a quadratic in x. Its discriminant decides everything: positive means two intersections, zero means the line is a tangent, negative means it misses. This one method answers all three cases.

What is the gradient of a line parallel to y = 4x − 7?

• A−7, which is the y-intercept rather than the gradient
• B−4, since parallel lines have gradients of the opposite sign
• C−1/4, which is the gradient of a line perpendicular to this one
• D4✓

Answer: Parallel lines have equal gradients, and in y = mx + c the gradient is m = 4. The perpendicular gradient (−1/4) and the intercept (−7) are both offered as traps.