Coordinate Geometry and Differentiation Test
1. 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]
2. The coordinates of two points are A (-1, -3) and B (5, 7). Calculate the equation of the
perpendicular bisector of AB.
[4]
3. 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.
[5]
(v) Calculate the coordinates of the point T where the line L meets SR.
[3]
(vi) Calculate the area of the quadrilateral PQRS.
[3]
4. Differentiate the following
(i) y  2 x(3x 2  5)
[1]
x  2x
x2
3  2x
(iii) y 
x
( x  2)(2 x  1)
(iv) y 
x4
5. A curve has equation y = x² – 3x + 1.
(i) Find the equation of the tangent to the curve at the point where x = 1.
(ii) Find the equation of the normal to the curve at the point where x = 3.
(ii)
y
3
6
[1]
[3]
[3]
[3]
[3]
6. A curve has equation y = 2x3  6x.
(iii) Show that the curve crosses the x-axis at the origin and the points ( 3 , 0) and
(  3 , 0)
dy
(iv) Find
. Hence find the stationary points on the curve.
dx
d2 y
(v) Find
, and use this to determine the nature of the stationary points.
dx 2
(vi) Sketch the curve.
7. Find the coordinates of the stationary points of the graph y 
the nature of each.
[2]
[3]
[3]
[2]
1 1 1
 
and determine
x x 2 x3
[5]
8. Two real numbers x and y are such that 2x + y = 100. Find the maximum value of the
product of the two numbers.
[5]
Total 60 marks
Solutions
1. (i)
5 y  4x  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
1  5 3  7 
The line passes through the midpoint of AB  
,
  (2, 2)
2 
 2
Equation of line is y  2   53 ( x  5 )
5( y  2)  3( x  2)
5 y  10  3 x  6
5 y  3 x  16
y 1  y 2 1  3 4 1



(i)
Gradient of PQ 
x 1  x 2 2  6 8 2
y1  y2 3  2
1
1



Gradient of QR 
x 1  x 2 6  9 3
3
y 1  y 2 2  ( 2) 4 1

 
Gradient of RS 
x1  x2
9 1
8 2
y 1  y 2 2  ( 1) 1
1



Gradient of SP 
x1  x2
1  ( 2)
3
3
(ii)
PQ is parallel to RS, and QR is parallel to SP, so the quadrilateral is a
parallelogram.
2. Gradient of AB 
3.
(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
y

15

2

7

1
When x = 7,
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
4. (i)
 16 5 5
 4  5  20
2
y  2 x(3 x  5 )  6 x 3  10 x
dy
dx
(ii)
y
dy
dx
(iii)
y
dy
dx
(iv)
y
dy
dx
 18 x 2  10
x3  2x6
 x  2x4
2
x
 1 8x3
3  2x
x
 3x 2  2 x 2
1
1
  23 x  2  x  2
3
1
( x  2)(2 x  1)
x
4
 2 x 2  3 x 3  2 x 4
 4 x 3  9 x 4  8 x 5
5. (i)
(ii)
y  x 2  3x  1
dy
 2x  3
dx
When x = 1, gradient  2  1  2  1
When x = 1, y  1 2  3  1  1
The tangent has gradient -1 and passes through (1, -1)
Equation of tangent is y  ( 1)  1( x  1)
y  1  x  1
y  x
When x = 3 , gradient of tangent is  2  3  3  3
Gradient of normal   31
When x = 3, y  3 2  3  3  1  1
Normal has gradient  31 and passes through (3, 1)
Equation of normal is y  1   31 ( x  3)
3( y  1)  ( x  3)
3y  3   x  3
3y  x  6
6. (i)
y  2 x 3  6x
When y = 0, 2 x 3  6 x  0
x 3  3x  0
x( x 2  3)  0
x  0 or x 2  3  x   3
So the graph cuts the x-axis at the origin and at the points

(ii)
3 ,0  .

dy
 6x 2  6
dx
At stationary points, 6 x 2  6  0
x2  1  0
( x  1)( x  1)  0
x  1 or x  1
When x = 1, y  2  1 3  6  1  2  6  4
When x = -1, y  2  1 3  6  1  2  6  4
so the stationary points are (1, -4) and (1, 4).
(iii)
d2 y
dx 2
 12 x
When x = 1,
When x = -1,
d2 y
dx 2
d2 y
dx 2
 12  0 so (1, -4) is a minimum point.
 12  0 so (-1, 4) is a maximum point.
3,0  and
(iv)
(-1, 4)
 3
3
(1, -4)
1
7. y 
dy
x

1
x
2

1
x3
 x 1  x 2  x 3
  x 2  2 x 3  3 x 4
dx
Stationary points occur where
dy
dx
0
0   x 2  2 x 3  3 x 4
3 x 4  x 2  2 x 3
3  x2  2x
x2  2x  3  0
( x  3)( x  1)  0
x  1 or x  3
When x = 1, y  1  1  1  1
When x = -3, y   31 
d2 y
dx 2
1
9
5
 271   27
 2 x 3  6 x 4  12 x 5
When x  3 :
d2 y
2
 2( 3) 3  6( 3)4  12( 3) 5
2
2
2
  27
 27
 814 
dx
d2 y
4
81
dx
5
As this is positive,  3,  27
 is a local minimum.
When x  1 :
d2 y
2
dx
d2 y
 2(1)3  6(1)4  12(1)5
 2  6  12  4
dx 2
As this is negative, (1,1) is a local maximum.
8. 2 x  y  100  y  100  2 x
Product of numbers P  xy  x(100  2 x )  100 x  2 x 2
dP
 100  4 x
dx
At stationary point, 100  4 x  0
x  25
P is a quadratic function, and the coefficient of x² is negative, so the stationary
point must be a maximum point.
When x = 25, y = 50, and P = 1250.
The maximum value of the product of x and y is 1250.