Pure Mathematics First Year Junior College

Test 4 Solutions

1) We have (2, 1), (4, 2), (1, 3) and (5, 1).

a) The gradient of line is 12

24=

.

The gradient of line is 2(3)

41=

.

The gradient of line is 3(1)

15=

.

The gradient of line is 1(1)

25=

.

Thus only and are parallel, and hence is a trapezium.

b) To divide internally in the ratio 3: 2, we convert the ratio to 1.5: 1 first, hence =

1.5. Then we substitute in the given formula.

= (1 + 2

1 + ,1 + 2

1 + ) = (

1 + 1.5(5)

1 + 1.5,3 + 1.5(1)

1 + 1.5) = (. , . )

c) The gradient of line is 1.81

3.42=

2.8

1.4= 2.

Since the product of the gradients of lines and is 2 1

2= 1, and are

perpendicular.

d) We use the formula =1

2( + ). is the length of line , is the length of line

and is the length of line .

Length of line is (2 3.4)2 + (1 (1.8))2

= (1.4)2 + 2.82 = 9.8 =7

55.

Length of line is (2 4)2 + (1 2)2 = (2)2 + (1)2 = 5.

Length of line is (1 5)2 + (3 (1))2

= (4)2 + (2)2 = 20 = 25.

Area of Trapezium =1

2(

7

55) (5 + 25) =

1

2(

7

55 35)

= . .

(Another way of working this problem out is to split the trapezium into two triangles and

finding the area of each triangle.)

2)

Distance from point (, ) to point (2, 3) is ( 2)2 + ( 3)2.

Distance from point (, ) to line 2 + 3 5 = 0 is

| + +

2 + 2| = |

2 + 3 5

22 + 32| =

1

13|2 + 3 5|

Therefore

( 2)2 + ( 3)2 =1

13|2 + 3 5|

Squaring both sides:

( 2)2 + ( 3)2 =(2 + 3 5)2

13

( 13) ( ) + ( ) = ( + )

3) The circle 2 + 2 + 10 + 6 66 = 0 has its centre at (10

2,

6

2) = (5, 3) and

radius = (5)2 + (3)2 (66) = 100 = 10.

The circle ( 10)2 + ( + 11)2 = 49 has its centre at (10, 11) and radius 49 = 7.

The distance between the two centres is (5 10)2 + (3 (11))2

=

(15)2 + 82 = 289 = 17

Since the sum of the two radii is equal to the distance between the two centres, the circles

touch externally.

The point where the two circles touch can be found by finding the point that divides

internally in the ratio 10: 7, where is the coordinates of the centre of the first circle and

is the coordinates of the centre of the second circle.

= (1 + 2

1 + ,1 + 2

1 + ) = (

5 + (107 )

(10)

1 +107

,3 + (

107 )

(11)

1 +107

) = ((

657 )

(177 )

,(

1317 )

(177 )

)

= (

,

).

4)

a) is a common radius for both circles, so the two circles must have the same radius.

Since and are also radii of each of the two circles, = = and hence triangle

is equilateral.

4b) Let be the midpoint of . Since is equilateral, lies vertically below and

hence its -coordinate is the same as that of , i.e. 1. Since lies horizontally to the right of

, its -coordinate is the same as that of , i.e. 1. Hence is at (1,1).

Since = and is the midpoint of , the distance is 1.5 times the distance .

But the distance is the distance from (2,1) to (1,1), which is equal to 1 (2) =

3 units, since is horizontal. Hence = 1.5 , which means that 3 = 1.5, and

hence = 2 units long. Hence the radii of both circles is 2 units long. Also, must be at

(0,1) and C must be at (2,1).

Thus the equation of the circle having centre at is ( 0)2 + ( 1)2 = 22, or +

( ) = . The equation of the circle having centre at is ( 2)2 + ( 1)2 = 22, or

( ) + ( ) = .