Straight Line Homework
Q 1 to 9
Total marks = 38
Q1 (a) Find the equation of the line joining
the points P(2 , -1) & Q(4 , -9)?
mPQ = -9 –(-1) = -9 + 1 = -8 = -4
4–2
2
2
(2, -1)
2
P
Q
If mPQ = -4 and passes thro P(2 , -1)
(4, -9)
y – (-1) = -4(x – 2)
y + 1 = -4x + 8
4x + y – 7 = 0
1
1
Q2 (a) Find altitude from A
Given A(-1 , 1); B(4 , 0) & C(1 , 6)?
A
mBC = 6 – 0 = 6 = -2
1 – 4 -3
If mBC = -2

1
mA = ½
1
Altitude from A(-1 , 1) with mA = ½ :y – 1 = ½ (x –(-1))
2y – 2 = x + 1
x - 2y + 3 = 0
2
C
B
Q3 The line y = 3x -1 makes an angle of ao
with the x-axis. Find the value of a.
Line y = 3x -1  m = 3
And
m = tanθ
tanθ = 3
m = tan -1(3)
m =71.565o
Therefore angle
ao
=
71.57o
(or
71.6o )
1
1
Q4 The diagonals of a polygon have equations
y = 2x + 1 and y = 3x – 2.
Find the point of intersection of the diagonals.
y = 2x + 1 and
y = 3x – 2
y=y
2x + 1 = 3x – 2
-x = -3
x =3
If y = 2x + 1 and x = 3 then:
y = 2(3) + 1
y=7
intersect when y = y
1
1
Thus point of
Intersection is at (3 , 7)
1, 1
Q5 A chord of a circle passes through the points
P(-2 , -3) and Q(6 , 1). Find the equation of the
diameter of the circle perpendicular to this chord
Midpoint of PQ, say R = (-2+6 , -3+1) = (2 , -1) 1
2
2
If mPQ = 1–(-3) = 4 = 1
6 – (-2) 8 2

mperp = -2
1
1
Perpendicular Bisector from PQ with mperp = -2:y – ( -1)= -2(x – 2)
y + 1 = -2x + 4
2x + y – 3 = 0
1
P
Q
Q6 P is the point (3 , 7) & Q is (7 , -1). Find
(a) Find the equation of the line PQ.
mPQ = 7 –(-1) = 8 = -2
3 – 7 -4
If mPQ = -2 and passes thro P(3 , 7)
1
P
1
Q
y – 7 = -2(x – 3)
y - 7 = -2x + 6
2x + y – 13 = 0
1
Q6 P is the point (3 , 7) & Q is (7 , -1). Find
(b) Find the equation of the line parallel to PQ,
through the point R(-1 , 2).
mPQ = 7 –(-1) = 8 = -2  mR = -2
3 – 7 -4
R
1
P
As parallel gradients are equal
If mR = -2 and passes thro R(-1 , 2)
y – 2 = -2(x – (-1))
y - 2 = -2x - 2
2x + y = 0 or y = -2x
1
1
Q
Q7 The curve y = 2x2 + x – 1 meets the line y = x + 1
at 2 points. Find the coordinates of both points.
If y = …. & y = …

y=y
2x2 + x – 1 = x + 1
2x2 = 2
x2 = 1
 x=±1
If x = -1 and y = x + 1:
y=x+1
= -1 + 1
=0
and
1
1
If x = 1 and y = x + 1:
y=x+1
=1+1
=2
Thus the two points of intersection are at (-1 , 0) and (1 , 2)
1
1
Q8 M is the point (a ,a2) , N is (3b , 9b2).
Find the gradient of MN in its simplest form.
1
1
1
m = (9b2 - a2 ) = (3b – a)(3b + a) = (3b + a)
(3b – a)
(3b – a)
Q9(a) In ΔLMN, L(-1 , -2), M(5 , 2) & N(1 , 8).
Find the equation of the median through N.
N
Median from N if L(-1 , -2) & M(5 , 2):Midpoint of LM, say K = (-1 + 5 , -2 + 2) = (2 , 0)
2
2
If mKN = 0 – 8
2–1
= -8 = -8
1
L
K
M
1
1
Median from N(1 , 8) with mKN = -8:y – 8 = -8 (x – 1)
y - 8 = -8x + 8
8x + y - 16 = 0
1
Q9(b) In ΔLMN, L(-1 , -2), M(5 , 2) & N(1 , 8).
Find the equation of the altitude through L.
L
mMN= 8 – 2 = 6 = -3
1 – 5 -4 2
If mMN = -3 
mL = ⅔
2
Altitude from L(-1 , -2) with mL = ⅔ :y – (-2) = ⅔(x – (-1))
3(y + 2) = 2(x + 1)
3y + 6 = 2x + 2
2x - 3y - 4 = 0
1
N
1
1
1
M
38 marks in total