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Coordinate geometry: AS Level Mathematics questions with answers Free Coordinate geometry questions for AS Level Mathematics (9709), part of Pure Mathematics 1. Each question has the answer and the marking points an examiner would look for.
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Coordinate geometry practice questions Question 1 Structured · 7 marks A* challenge
The curve C has equation y = x2 − 6x + 6. The point P on C has x-coordinate 4.
(a) Find the equation of the normal to C at P.
[3] (b) The normal meets C again at Q. Find the x-coordinate of Q.
[4] Show answers and mark scheme Part Answer and marking points Marks (a) y = −1 2 dy dx B1 1 2 M1 A1
3 (b) 3 2 substitute normal into curve M1 A1 M1 3 2 A1
4
Question 2 Structured · 10 marks A* challenge
The circle C has equation (x)2 + (y − 2)2 = 25. The line L has equation x + 2y = 14.
(a) Find the coordinates of the points P and Q where L meets C.
[4] (b) Find the equation of the tangent to C at the point (0, 7).
[3] (c) The tangents to C at P and Q meet at the point T. Find the coordinates of T.
[3] Show answers and mark scheme Part Answer and marking points Marks (a) (0, 7) and (4, 5)substitute line into circle M1 A1 M1 A1
4 (b) y = 7radius is vertical, so the tangent is horizontal B1 M1 A1
3 (c) (5 2 equation of second tangent B1 M1 5 2 A1
3
Question 3 Structured · 6 marks
(a) A circle has centre C(−1, −4) and passes through the point P(4, 8). Find the equation of the circle.
[3] (b) Find the equation of the tangent to the circle at P.
[3] Show answers and mark scheme Part Answer and marking points Marks (a) (x + 1)2 + (y + 4)2 = 169Radius2 by Pythagoras M1 2 = 169 A1 A1
3 (b) y − 8 = −5 12 gradient CP = 12 5 M1 5 12 M1 A1
3
Question 4 Structured · 8 marks A* challenge
The points A and B have coordinates (1, −2) and (5, −4).
(a) Find the equation of the perpendicular bisector of AB.
[4] (b) The point C lies on the perpendicular bisector and has x-coordinate 5. Find the y-coordinate of C.
[1] (c) Find the area of triangle ABC.
[3] Show answers and mark scheme Part Answer and marking points Marks (a) y + 3 = 2(x − 3)mid-point (3, −3) B1 1 2 B1 M1 A1
4 (b) 1y = 1 B1
1 (c) 10AB and height MC found (or shoelace) M1 A1 A1
3
Question 5 Structured · 9 marks A* challenge
A circle has equation x2 + y2 − 8x − 9 = 0.
(a) Find the centre and radius of the circle.
[3] (b) Show that the point P (0, −3) lies on the circle.
[1] (c) Find the equation of the tangent to the circle at P.
[3] (d) The tangent meets the x-axis at T. Find the length PT.
[2] Show answers and mark scheme Part Answer and marking points Marks (a) centre (4, 0), radius 5completing the square in x and y M1 A1 A1
3 (b) substitution gives 0substitute and obtain 0 (or distance from centre = 5) B1
1 (c) y + 3 = −4 3 gradient of radius 3 4 B1 4 3 M1 4 3 A1
3 (d) PT = 3.75x-coordinate of T = −9 4 M1 A1
2
Question 6 Structured · 9 marks A* challenge
The points A (1, −2), B (3, −1) and D (0, 0) are three vertices of a parallelogram ABCD.
(a) Find the coordinates of C.
[2] (b) Show that ABCD is a rectangle.
[3] (d) Find the equation of the diagonal AC.
[2] Show answers and mark scheme Part Answer and marking points Marks (a) (2, 1)AB vector added to D, or mid-points of diagonals equal M1 A1
2 (b) gradients multiply to −1gradient AB = 1 2 B1 B1 B1
3 (c) 5AB × AD M1 A1
2 (d) y + 2 = 3(x − 1)gradient of AC M1 A1
2
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