EdExcel C1 Coordinate geometry Assessment solutions

Coordinate Geometry

Chapter assessment

  1. Find the coordinates of the points where the line 5y + 2x + 10 = 0 meets the axes and hence sketch the line. [2]
  2. A line l1 has equation .

(i)Find the gradient of the line.[1]

(ii)Find the equation of the line l2 which is parallel to l1 and passes through the point
(1, -2).[3]

  1. The coordinates of two points are A (-1, -3) and B (5, 7).

(i)Find the equation of the line l which is perpendicular to AB and passes through the point B. [4]

(ii)The line l crosses the y-axis at the point C.
Find the area of triangle ABC, simplifying your answer as far as possible.[4]

  1. The line meets the x-axis at the point P, and the line meets the x-axis at the point Q.
    The two lines intersect at the point R.

(i)Find the coordinates of R.[4]

(ii)Find the area of triangle PQR.[3]

  1. The coordinates of four points are P (-2, -1), Q (6, 3), R (9, 2) and S (1, -2).

(i)Calculate the gradients of the lines PQ, QR, RS and SP.[4]

(ii)What name is given to the quadrilateral PQRS?[1]

(iii)Calculate the length SR.[2]

(iv)Show that the equation of SR is 2y = x – 5 and find the equation of the line L through
Q perpendicular to SR.[6]

(v)Calculate the coordinates of the point T where the line L meets SR.[3]

(vi)Calculate the area of the quadrilateral PQRS.[3]

Total 40 marks

Coordinate Geometry

Solutions to Chapter assessment


  1. When x = 0,
    When y = 0,
    The line meets the axes at (0, -2) and (-5, 0).

  2. (i).
    Gradient of line =
    (ii)l2 is parallel to l1, so it has gradient .
    Equation of line is
  3. (i)Gradient of AB
    Gradient of line perpendicular to AB .
    Equation of line is
    (ii)
    When x = 0, y = 10
    The coordinates of C are (0, 10)
    Triangle ABC has a right angle at B.
    LengthAB
    LengthBC
    Area
  4. (i)Substituting into :

    When x = 1.7,
    The coordinates of R are (1.7, 0.4)
    (ii)P is the point on where y = 0, so P is (1.5, 0)
    Q is the point on where y = 0, so Q is (2, 0).
    Area
  5. (i)Gradient of PQ
    Gradient of QR
    Gradient of RS
    Gradient of SP
    (ii)PQ is parallel to RS, and QR is parallel to SP, so the quadrilateral is a
    parallelogram.
    (iii)
    (iv) From (i), gradient of SR
    Equation of SR is
    Line perpendicular to SR has gradient -2
    Line L has gradient -2 and goes through (6, 3)
    Equation of L is
    (v)Equation of L is
    Substituting into equation of SR gives
    When x = 7,
    Coordinates of T are (7, 1)
    (vi)

    Length QT
    Area of parallelogram

© MEI, 09/07/121/4