Cambridge International Mathematics

Straight lines (Chapter 14) 309

Example 10 Self Tutor

ABC is an isosceles triangle with AB=AC.

A(0,¡1),B( 5 , 1 ) and C(2,k) are the coordinates of A, B and

C, with k> 0.

a Explain why k=4.

b Find the coordinates of M, the midpoint of line segment BC.

c Show that line segments AM and BC are perpendicular.

d Find the equation of the line of symmetry of triangle ABC.

a

b

c

d

EXERCISE 14F

1aPlot the points A(1,0),B(9,0),C(8,3) and D(2,3).

b Classify the figure ABCD.

c Find the equations of any lines of symmetry of ABCD.

2aPlot the points A(¡ 1 ,2),B(3,2),C(3,¡2) and D(¡ 1 ,¡2).

b Classify the figure ABCD.

c Find the equations of any lines of symmetry of ABCD.

3 P, Q, R and S are the vertices of a rectangle. Given P(1,3),Q(7,3)and S( 1 ,¡2) find:

a the coordinates of R

b the midpoints of line segments PR and QS

c the equations of the lines of symmetry.

d What geometrical fact was verified byb?

O

y

x

A,()0 -1

B,()5 ¡1

C,()2 ¡k

M

Since AB=AC,

p

(5¡0)^2 +(1¡¡1)^2 =

p

(2¡0)^2 +(k¡¡1)^2

)

p

25 + 4 =

p

4+(k+1)^2

) (k+1)^2 =25

) k+1=§ 5

) k=4or ¡ 6

But k> 0 ,sok=4.

The midpoint of BC is M

¡5+2

2 ,

1+4

2

¢

.

So, M is

¡ 7

2 ,

5

2

¢

:

gradient of BC=

4 ¡ 1

2 ¡ 5

=¡^33

=¡ 1

gradient of AM=

5

2 ¡¡^1

7

2 ¡^0

=

7

2

7

2

=1

As these gradients are

negative reciprocals of

each other, AM?BC.

The line of symmetry is AM.

Its gradient m=1and itsy-intercept c=¡ 1.

) its equation is y=x¡ 1.

IGCSE01

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Y:\HAESE\IGCSE01\IG01_14\309IGCSE01_14.CDR Friday, 17 October 2008 9:50:12 AM PETER