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PQ applied <strong>to</strong> T 3<br />

give ( )( ) = ( ) and this is a reflection in the line y = – x.<br />

The inverse of a reflection in the line y = x is itself a reflection in the line y = x.<br />

7 (a) Reflection in the line x + y = 0, i.e. y = – x<br />

(Hint: if in doubt, join corresponding points on each triangle and bisect these lines.)<br />

(b) A = (<br />

3 4 3<br />

) B = ( – 2 2 5<br />

2 2 5<br />

)<br />

– 3 – 4 – 3<br />

a b 3 4 3<br />

( )( ) = ( – 2 2 5<br />

c d<br />

)<br />

2 2 5 – 3 – 4 – 3<br />

3a + 2b 4a + 2b 3a + 5b<br />

( ) = ( – 2 2 5<br />

)<br />

3c + 2d 4c + 2d 3c + 5d – 3 – 4 – 3<br />

Comparing elements:<br />

3a + 2b = 2<br />

4a + 2b = 2<br />

so a = 0 and b = 1<br />

Also 3c + 2d = – 3<br />

4c + 2d = – 4<br />

so c = – 1 and d = 0<br />

Hence the transformation matrix = (<br />

0 1<br />

)<br />

– 1 0<br />

Model answers<br />

(c) By comparing corresponding lengths, the scale fac<strong>to</strong>r is – 2. By joining corresponding points with straight lines<br />

and finding the intersection of these lines, the centre is at (1, 5).<br />

2 1 3 4 3<br />

(d) (i) ( 0 3)( 2 2 5) = ( 6 6 15)<br />

1 1<br />

(ii) Area = (10 – 8) × – (15 – 6) = –<br />

2<br />

2 × 2 × 9 = 9 square units<br />

(iii) (<br />

a b<br />

)(<br />

8 10 11<br />

) = (<br />

3 4 3<br />

c d 6 6 15 2 2 5 )<br />

8a + 6b 10a + 6b 11a + 15b<br />

( ) = 3 4 3<br />

( )<br />

8c + 6d 10c + 6d 11c + 15d 2 2 5<br />

Comparing elements:<br />

8a + 6b = 3<br />

10a + 6b = 4<br />

1<br />

so 2a = 1, thus a = –<br />

2<br />

3 – 4<br />

and b = =<br />

– 1<br />

–––––– ––6<br />

6<br />

Also 8c + 6d = 2<br />

10c + 6d = 2<br />

1<br />

so c = 0 = d = –<br />

3<br />

8 (a) (i) A shear along the x-axis direction, shear fac<strong>to</strong>r = – 2 (or a shear with the x-axis as the invariant line)<br />

(ii) A shear along the y-axis direction, shear fac<strong>to</strong>r = 2 (or a shear with the y-axis as the invariant line)<br />

(b) (<br />

a b<br />

)(<br />

0 3 5 2<br />

) = (<br />

0 3 3 0<br />

c d 0 0 1 1 0 0 2 2 )<br />

3a = 3 thus a = 1<br />

5a + b = 3 thus b = – 2<br />

Also<br />

5c + d = 2 thus c = 0<br />

2c + d = 2 thus d = 2<br />

so the matrix is (<br />

1<br />

– 2<br />

0 2 )<br />

9 (a) (i) A par<strong>all</strong>elogram<br />

0<br />

– 1<br />

– 2 3 4<br />

– 1 0<br />

– 4 – 8 – 6<br />

1 –2<br />

8 10 11<br />

1 1 1 3 3 1<br />

( )( ) = 2 4 7 5<br />

( )<br />

0 1 1 1 4 4 1 1 4 4<br />

(ii) A shear with the x-axis as the invariant line and shear fac<strong>to</strong>r 1<br />

(b) (i) The original rectangle<br />

1<br />

( – 1 2 4 7 5<br />

)( ) = 1 3 3 1<br />

0 1<br />

( )<br />

1 1 4 4 1 1 4 4<br />

– 4 8 6<br />

– 2 – 3 – 4<br />

(ii) A shear with the x-axis as the invariant line and shear fac<strong>to</strong>r = – 1<br />

20<br />

IGCSE Revision Guide for Mathematics Model <strong>Answers</strong> © 2004, <strong>Hodder</strong> & S<strong>to</strong>ugh<strong>to</strong>n Educational

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