Higher Homework 1
The Straight Line
Total marks = 28
Q1 A Quadrilateral has vertices
A(–1, 8),B(7, 12), C(8, 5) & D(2, –3)as shown
B(7, 12)
y
A(– 1, 8)
E
C(8, 5)
x
O
D(2, – 3)
Q1 (a) Find the equation of the diagonal BD.
B(7, 12)
D(2, –3)
mBD= 12 –(-3) = 15 = 3
7–2
1
2
B(7, 12)
5
y
A(– 1, 8)
E
If mBD = 3 & passes thro B(7, 12)
C(8, 5)
(y – 12) = 3(x – 7)
y – 12 = 3x – 21
y = 3x – 9
or
3x – y – 9 = 0
x
O
D(2, – 3)
1
2
Q1 (a) Alternatively use point D:
B(7, 12)
D(2, –3)
mBD= 12 –(-3) = 15 = 3
7–2
1
B(7, 12)
5
y
A(– 1, 8)
E
If mBD = 3 & passes thro D(2, -3)
C(8, 5)
(y – (-3)) = 3(x – 2)
y + 3 = 3x – 6
y = 3x – 9
or
3x – y – 9 = 0
x
O
D(2, – 3)
1
2
Q1 (b) The equation AC is x + 3y = 23. Find the
co-ordinates of E where AC & BD intersect.
3x – y = 9
x + 3y = 23
1
If x = 5 subst into (2):
2
x + 3y = 23
5 + 3y = 23
(1) x 3: 9x – 3y = 27
x + 3y = 23
3
3y = 18
1
1
y = 6  E( 5 , 6)
10x = 50
x=5
1
3
Q1 (c) (i) Find the equation the perpendicular bisector of AB.
(ii) Show the line passes thro E.
5
A(–1, 8) mAB= 12 – 8 = 4 = ½  mPERP = -2 (As m1 x m2 = -1)
1
1
7 – (-1) 8
B(7, 12)
B(7, 12)
M
y
Biscects  Midpoint
8+12
6 20
(–1+7
,
)
=
(
, 2 ) = ( 3 , 10 )
2
2
2
A(– 1, 8)
1
C(8, 5)
If mPERP = -2 & passes thro M(3,10)
(y – 10) = -2(x – 3)
y – 10 = -2x + 6
y = -2x + 16
or 2x + y – 16 = 0
E(5, 6)
x
O
D(2, – 3)
1
4
Q1 (c)
E(5, 6)
(ii) Show the line passes thro E.
1
Perpendicular bisector AB: y = -2x + 16
M
If x = 5: y = -2x + 16
y = -2(5) + 16
y = -10 + 16
y=6
E(5, 6) fits
B(7, 12)
y
A(– 1, 8)
E(5, 6)
C(8, 5)
1
x
O
D(2, – 3)
E lies on perpendicular bisector of AB
1
Q2 (a) State the coordinates of P
1
If P cuts the x-axis  y = 0
Q(4, 6)
6x – 7y + 18 = 0
6x – 0 + 18 = 0
6x = –18
 x = –3
 P(– 3, 0)
y
T
6x – 7y + 18 = 0
O
1
x
P
R(8, –2)
1
Q2 (b) Find the equation of the altitude of the triangle from P 3
Q(4, 6) mQR= 6 – (-2) = 8 = -2  mPERP = ½ (As m1 x m2 = -1)
1
1
4–8
-4
R(8, –2)
Q(4, 6)
If mPERP = ½ & passes thro P(-3,0)
y
T
6x – 7y + 18 = 0
(y – 0) = ½ (x – (-3))
2(y) = x + 3
O
x
P(– 3, 0)
x - 2y + 3 = 0
R(8, –2)
1
Equation of Altitude from P is: x – 2y = -3
3
Q2 (c) The altitude from P meets the line QR at T.
Find the co-ordinates of T.
4
Q(4, 6) mQR= 6 – (-2) = 8 = -2 (already found for (a))
4–8
-4
R(8, –2)
Q(4, 6)
y
(y – 6) = -2 (x – 4)
6x – 7y + 18 = 0
T
y – 6 = –2x + 8
O
y = –2x + 14
or
2x + y – 14 = 0
x
P(– 3, 0)
1
R(8, –2)
Equation QR: 2x + y = 14
1
Q2 (c) The altitude from P meets the line QR at T.
Find the co-ordinates of T.
1
x – 2y = -3
y
Q(4, 6)
2 1 6x – 7y + 18 = 0
2x + y = 14
R(8, –2)
x – 2y = -3
Q(4, 6)
T
O
x
P(– 3, 0)
(2) x 2: 4x + 2y = 28
R(8, –2)
5x = 25
 x=5
4
1
If x = 5 subst into (2):
2x + y = 14
2(5) + y = 14
1
10 + y = 14
y = 4  T( 5 , 4)
3
Q3. The vertices of triangle ABC are A(7, 9), B(-3,-1) and C(5, -5).
The broken line represents the perpendicular bisector of BC.
A(7, 9)
y
x
O
B(–3, –1)
M
C(5, –5)
Q3 (a) Show the perpendicular bisector of BC is y = 2x – 5
4
mBC = -1 – (-5) = 4 = -½  mPERP = 2 (As m1 x m2 = -1)
A(7, 9)
-3 – 5
-8
A(7, 9)
1
B(-3, -1)
C(5, –5) Biscects  Midpoint
( –3+5 , -1+(-5) ) = ( 22 , -62 ) = ( 1 , -3 )
2
1
y
2
If mPERP = -2 & passes thro M(1, -3)1
(y – (-3)) = 2(x – 1)
B(–3, –1)
y + 3 = 2x – 2
 y = 2x – 5 As required
x
O
M
C(5, –5)
1
4
Q3 (b) Find the equation of the median from C.
3
Median  Midpoint
A(7, 9)
–3 + 7 ,
(
B(-3, -1)
2
-1 + 9
2
) = ( 42 ,
8
2
A(7, 9)
)  M(2, 4)
1
C(5, –5) mCM = 4 – (-5) = 9 = -3 1
2–5
-3
M(2, 4)
If mCM = -2 & passes thro C(5, -5)
y
x
O
(y – (-5)) = -3(x – 5)
y + 5 = -3x + 15
 y = -3x + 10
or
or
B(–3, –1)
M
C(5, –5)
1
3x + y – 10 = 0
3x + y = 10
3
Q3 (b) Find the where the two lines intersect
3x + y = 10
2x - y = 5
5x = 15
 x=3
3
1
2
1
A(7, 9)
y
1
If x = 3 subst into (1):
3x + y = 10
3( 3 ) + y = 10
9 + y = 10
y = 1  ( 3 , 1)
x
O
B(–3, –1)
M
C(5, –5)
1
3
28 marks in total