```EdExcel Core 1
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 5 y  4 x  3 .
(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]
3. 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]
4. The line y  2 x  3 meets the x-axis at the point P, and the line 3 y  4 x  8 meets the xaxis 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]
5. 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
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EdExcel C1 Coordinate geometry Assessment solutions
Coordinate Geometry
Solutions to Chapter assessment
1. 5 y  2 x  10  0
When x = 0, 5 y  10  0  y  2
When y = 0, 2 x  10  0  x  5
The line meets the axes at (0, -2) and (-5, 0).
-5
-2
2. (i) 5 y  4 x  3 .
5 y  4 x  3
y   54 x  53
Gradient of line =  54
(ii) l2 is parallel to l1, so it has gradient  54 .
Equation of line is y  ( 2)   54 ( x  1)
5( y  2)  4( x  1)
5 y  10  4 x  4
5 y  4x  6  0
y 1  y 2 3  7 10 5



x 1  x 2 1  5
6
3
3
Gradient of line perpendicular to AB   .
5
Equation of line is y  7   53 ( x  5 )
5( y  7)  3( x  5 )
5 y  35  3 x  15
5 y  3 x  50
3. (i) Gradient of AB 
(ii) 5 y  3 x  50
When x = 0, y = 10
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EdExcel C1 Coordinate geometry Assessment solutions
The coordinates of C are (0, 10)
Triangle ABC has a right angle at B.
C (0, 10)
Length AB  (5  ( 1))2  ( 7  ( 3))2
B (5, 7)
 36  100
 136
Length BC  (5  0)2  ( 7  10)2
 25  9
A (-1, -3)
 34
Area  21  AB  BC

1
2
136  34

1
2
4 34 34
 21  2  34
 34
4. (i) Substituting y  2 x  3 into 3 y  4 x  8 :
3(2 x  3)  4 x  8
6x  9  4x  8
10 x  17
x  1.7
When x = 1.7, y  2  1.7  3  3.4  3  0.4
The coordinates of R are (1.7, 0.4)
R (1.7, 0.4)
(ii) P is the point on y  2 x  3 where y = 0, so P is (1.5, 0)
Q is the point on 3 y  4 x  8 where y = 0, so Q is (2, 0).
Area  21  base  height
 21  0.5  0.4
P (1.5, 0)
Q (2, 0)
 0.1
y 1  y 2 1  3 4 1



x 1  x 2 2  6 8 2
y1  y2 3  2
1
1



x 1  x 2 6  9 3
3
y 1  y 2 2  ( 2) 4 1

 
x1  x2
9 1
8 2
y 1  y 2 2  ( 1) 1
1



x1  x2
1  ( 2)
3
3
5. (i) Gradient of PQ 
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EdExcel C1 Coordinate geometry Assessment solutions
(ii) PQ is parallel to RS, and QR is parallel to SP, so the quadrilateral is a
parallelogram.
(iii) SR  ( 9  1)2  (2  ( 2))2  64  16  80
(iv) From (i), gradient of SR  21
Equation of SR is y  ( 2)  21 ( x  1)
2( y  2)  x  1
2y  4  x  1
2y  x  5
Line perpendicular to SR has gradient -2
Line L has gradient -2 and goes through (6, 3)
Equation of L is y  3  2( x  6)
y  3  2 x  12
y  2 x  15
(v) Equation of L is y  15  2 x
Substituting into equation of SR gives 2(15  2 x )  x  5
30  4 x  x  5
35  5 x
x7
When x = 7, y  15  2  7  1
Coordinates of T are (7, 1)
(vi)
L
Q (6, 3)
R (9, 2)
T (7, 1)
P
(-2, -1)
S (1, -2)
Length QT  ( 7  6)2  (1  3)2  1  4  5
Area of parallelogram  SR  QT
 80 5
 16 5 5
 4  5  20
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