MT_F4_7.2 Parallel & Perpendicular Lines_2

Solving problems involving equations of parallel and perpendicular lines

1) The diagram shows a triangle ABC, such that A (9, 9) and C (1, –3). Point
B lies on the perpendicular bisector of AC and the equation of straight line

AB is y  8x  63.

(a) Find 900 两个相等部分

(i) the equation of perpendicular bisector y perpendicular bisector

of AC, A (9, 9)

(ii) the coordinates of B.

(b) Point D lies on the diagram such that y = 8x – 63

ABCD is a rhombus. B x
(i) Find the coordinates of D.
(ii) Show that AC = 2BD. 0

C (1, –3)

(a) (i) mAC  9   3 m1m2  1 y

9 1 A (9, 9)

3 m2   2 y = 8x – 63
2 3
B
 9 1, 9   3 x
0
Midpoint AC = 2 2 
C (1, –3)

y  mx  c  5, 3

Gradient form The equation of perpendicular bisector AC is

3    2 5 c y   2 x  19
33
 3

c  19
3

Or The equation of perpendicular bisector AC is

y  y1  mx  x1 y  3   2 x  5

General form 3
3y  9  2x 10

2x  3y 19  0

(ii) y   2 x  19 1 y
33 2
A (9, 9)
y  8x  63

 2 x  19  8x  63 y = 8x – 63
33
2x 19  24x 189 0 B

26x  208 C (1, –3) x
x8
Intersection point
When x = 8, y  88 63
 B8, 1
1

 (b) (i) Midpoint AC = Midpoint BD  5, 3 y

A (9, 9)

 x8, y  1   5, 3 D 5, 3
2 2

x8 5 y 1  3 B x
2 2
0

C (1, –3)

x  8 10 y 1 6

x2 y5 ** ABCD is a rhombus

 D2, 5 - Two pairs of parallel lines
- Intersection point = midpoint
- All lengths are equal

** Parallelogram
- Two pairs of parallel lines
- Intersection point = midpoint

(ii) A9, 9 C1,3 Distance

AC  9   32  9 12 y2  y12  x2  x12

 208
 4 13

B8, 1 D2, 5

BD  1 52  8  22

 52
 2 13

AC  4 13
BD 2 13
AC  2
BD
 AC  2BD (shown)

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