7
Linear Functions
TERMINOLOGY
Collinear points: Two or more points that lie on the same
straight line
Interval: A section of a straight line including the end
points
Concurrent lines: Two or more lines that intersect at a
single point
Midpoint: A point lying exactly halfway between two
points
Gradient: The slope of a line measured by comparing
the vertical rise over the horizontal run. The symbol for
gradient is m
Perpendicular distance: The shortest distance between a
point and a line. The distance will be at right angles to
the line
Chapter 7 Linear Functions
INTRODUCTION
IN CHAPTER 5, YOU STUDIED functions and their graphs. This chapter
looks at the linear function, or straight-line graph, in more detail.
Here you will study the gradient and equation of a straight line, the
intersection of two or more lines, parallel and perpendicular lines,
the midpoint, distance and the perpendicular distance from a point
to a line.
DID YOU KNOW?
Pierre de Fermat (1601–65) was a lawyer who dabbled in mathematics. He was a contemporary of
Descartes, and showed the relationship between an equation in the form Dx = By, where D and
B are constants, and a straight-line graph. Both de Fermat and Descartes only used positive values
of x, but de Fermat used the x-axis and y-axis as perpendicular lines as we do today.
De Fermat’s notes Introduction to Loci, Method of Finding Maxima and Minima and Varia
opera mathematica were only published after his death. This means that in his lifetime de Fermat
was not considered a great mathematician. However, now he is said to have contributed as
much as Descartes towards the discovery of coordinate geometry. De Fermat also made a great
contribution in his discovery of differential calculus.
Class Assignment
Find as many examples as you can of straight-line graphs in newspapers
and magazines.
Distance
The distance between two points (or the length of the interval between two
points) is easy to find when the points form a vertical or horizontal line.
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EXAMPLES
Find the distance between
1. ^ -1, 4 h and ^ -1, -2 h
Solution
Counting along the y-axis, the distance is 6 units.
2. ^ 3, 2 h and ^ -4, 2 h
Solution
Counting along the x-axis, the distance is 7 units.
When the two points are not lined up horizontally or vertically, we use
Pythagoras’ theorem to find the distance.
Chapter 7 Linear Functions
393
EXAMPLE
Find the distance between points ^ 3, -1 h and ^ -2, 5 h.
Solution
BC = 5 and AC = 6
By Pythagoras’ theorem,
You studied Pythagoras’
theorem in Chapter 4.
c =a +b
AB 2 = 5 2 + 6 2
= 25 + 36
= 61
2
2
2
` AB = 61
Z 7.81
DID YOU KNOW?
Pythagoras made many discoveries about music as well as about mathematics. He found
that changing the length of a vibrating string causes the tone of the music to change. For
example, when a string is halved, the tone is one octave higher.
The distance between two points _ x 1, y 1 i and _ x 2, y 2 i is given by
d=
2
2
_ x2 - x1 i + _ y2 - y1 i
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Proof
If points A and B were changed
around, the formula would be
d =
(x 1 - x 2 ) + (y 1 - y 2 ) ,
2
2
which would give the same
answer.
Let A = _ x 1, y 1 i and B = _ x 2, y 2 i
Length AC = x 2 - x 1 and length BC = y 2 - y 1
By Pythagoras’ theorem
AB 2 = AC 2 + BC 2
d 2 = _ x 2 - x 1 i2 + _ y 2 - y 1 i2
`d=
2
2
_ x2 - x1 i + _ y2 - y1 i
EXAMPLES
1. Find the distance between the points ^ 1, 3 h and ^ -3, 0 h.
Solution
Let ^ 1, 3 h be _ x 1, y 1 i and ^ -3, 0 h be _ x 2, y 2 i
d=
2
2
_ x2 - x1 i + _ y2 - y1 i
= ] -3 - 1 g2 + ] 0 - 3 g2
= ] -4 g2 + ] -3 g2
= 16 + 9
= 25
=5
So the distance is 5 units.
2. Find the exact length of AB given that A = ^ -2, -4 h and B = ^ -1, 5 h .
Solution
Let ^ -2, -4 h be _ x 1, y 1 i and ^ -1, 5 h be _ x 2, y 2 i
d=
You would still get 82 if you
used (- 2, - 4) as (x 2 , y 2 ) and
(-1, 5) as (x 1 , y 1 ).
2
2
_ x2 - x1 i + _ y2 - y1 i
=
6 -1 - ^ -2 h @ 2 + 6 5 - ^ -4 h @ 2
=
=
=
12 + 92
1 + 81
82
Chapter 7 Linear Functions
7.1 Exercises
1.
Find the distance between points
(a) ^ 0, 2 h and ^ 3, 6 h
(b) ^ -2, 3 h and ^ 4, -5 h
(c) ^ 2, -5 h and ^ -3, 7 h
2.
Find the exact length of the
interval between points
(a) ^ 2, 3 h and ^ -1, 1 h
(b) ^ -5, 1 h and ^ 3, 0 h
(c) ^ - 2, -3 h and ^ - 4, 6 h
(d) ^ -1, 3 h and ^ -7, 7 h
3.
4.
Find the distance, correct to
2 decimal places, between points
(a) ^ 1, -4 h and ^ 5, 5 h
(b) ^ 0, 4 h and ^ 3, -2 h
(c) ^ 8, -1 h and ^ -7, 6 h
Find the perimeter of D ABC with
vertices A ^ 3, 1 h, B ^ -1, 1 h and
C ^ -1, -2 h .
5.
Prove that the triangle with
vertices ^ 3, 4 h, ^ -2, 7 h and ^ 6, -1 h
is isosceles.
6.
Show that AB = BC, where
A = ^ -2, 5 h, B = ^ 4, -2 h and
C = ^ -3, -8 h .
7.
8.
9.
Show that points ^ 3, -4 h and ^ 8,1 h
are equidistant from point ^ 7, -3 h .
A circle with centre at the origin
O passes through the point
_ 2 , 7 i . Find the radius of the
circle, and hence its equation.
Prove that the points
X _ 2 , -3 i, Y _ -1, 10 i and
Z _ - 6 , 5 i all lie on a circle
with centre at the origin. Find its
equation.
10. If the distance between ^ a, -1 h
and ^ 3, 4 h is 5, find the value of a.
11. If the distance between ^ 3, -2 h
and ^ 4, a h is 7 , find the exact
value of a.
12. Prove that A ^ 1, 4 h, B ^ 1, 2 h and
C _ 1 + 3 , 3 i are the vertices of
an equilateral triangle.
13. If the distance between ^ a, 3 h
and ^ 4, 2 h is 37 , find the values
of a.
14. The points M ^ -1, -2 h, N (3, 0),
P ^ 4, 6 h and Q ^ 0, 4 h form
a quadrilateral. Prove that
MQ = NP and QP = MN. What
type of quadrilateral is MNPQ?
15. Show that the diagonals
of a square with vertices
A ^ -2, 4 h, B ^ 5, 4 h, C ^ 5, -3 h and
D ^ -2, -3 h are equal.
16. (a) Show that the triangle with
vertices A ^ 0, 6 h, B ^ 2, 0 h and
C ^ -2, 0 h is isosceles.
(b) Show that perpendicular OA,
where O is the origin, bisects BC.
17. Find the exact length of the
diameter of a circle with centre
^ -3, 4 h if the circle passes
through the point ^ 7, 5 h .
18. Find the exact length of the
radius of the circle with centre
(1, 3) if the circle passes through
the point ^ -5, -2 h .
19. Show that the triangle
with vertices A ^ -2, 1 h, B ^ 3, 3 h
and C ^ 7, -7 h is right angled.
20. Show that the points
X ^ 3, -3 h, Y ^ 7, 4 h and Z ^ - 4, 1 h
form the vertices of an isosceles
right-angled triangle.
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Midpoint
The midpoint is the point halfway between two other points.
The midpoint of two points _ x 1, y 1 i and _ x 2, y 2 i is given by
M=e
x1 + x2 y1 + y2
o
,
2
2
Proof
Can you see why
these triangles
are similar?
Find the midpoint of points A _ x 1, y 1 i and B _ x 2, y 2 i.
Let M = ^ x, y h
Then D APQ <; D ABR
AQ
AP
=
AR
AB
x - x1
1
` x -x =
2
2
1
2 _ x - x1 i = x2 - x1
2x - 2x 1 = x 2 - x 1
2x = x 1 + x 2
x1 + x2
`
x=
2
y1 + y2
Similarly, y =
2
`
EXAMPLES
1. Find the midpoint of ^ -1, 4 h and ^ 5, 2 h.
Solution
x=
x1 + x2
2
Chapter 7 Linear Functions
-1 + 5
2
4
=
2
=2
y1 + y2
y=
2
4+2
=
2
6
=
2
=3
So M = (2, 3) .
=
2. Find the values of a and b if ^ 2, -3 h is the midpoint between ^ -7, -8 h
and ^ a, b h.
Solution
x=
x1 + x2
2
-7 + a
2=
2
4 = -7 + a
11 = a
y1 + y2
y=
2
-8 + b
-3 =
2
-6 = -8 + b
2=b
So a = 11 and b = 2.
Note that the x-coordinate of the midpoint is the average of x 1 and x 2 .
The same applies to the y-coordinate.
PROBLEM
A timekeeper worked out the average time for 8 finalists in a race. The
average was 30.55, but the timekeeper lost one of the finalist’s times.
The other 7 times were 30.3, 31.1, 30.9, 30.7, 29.9, 31.0 and 30.3.
Can you find out the missing time?
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7.2 Exercises
1.
2.
The locus is the path
that P (x, y) follows.
Find the midpoint of
(a) ^ 0, 2 h and ^ 4, 6 h
(b) ^ -2, 3 h and ^ 4, -5 h
(c) ^ 2, -5 h and ^ -6, 7 h
(d) ^ 2, 3 h and ^ -8, 1 h
(e) ^ -5, 2 h and ^ 3, 0 h
(f) ^ -2, -2 h and ^ -4, 6 h
(g) ^ 1, -4 h and ^ 5, 5 h
(h) ^ 0, 4 h and ^ 3, -2 h
(i) ^ 8, -1 h and ^ -7, 6 h
(j) ^ 3, 7 h and ^ -3, 4 h
Find the values of a and b if
(a) ^ 4, 1 h is the midpoint of ^ a, b h
and ^ -1, 5 h
(b) ^ -1, 0 h is the midpoint of
^ a, b h and ^ 3, -6 h
(c) ^ a, 2 h is the midpoint of (3, b h
and ^ -5, 6 h
(d) ^ -2, 1 h is the midpoint of
^ a, 4 h and ^ -3, b h
(e) ^ 3, b h is the midpoint of ^ a, 2 h
and ^ 0, 0 h
3.
Prove that the origin is the
midpoint of ^ 3, -4 h and ^ -3, 4 h .
4.
Show that P = Q where P is the
midpoint of ^ -2, 3 h and ^ 6, -5 h
and Q is the midpoint of ^ -7, -5 h
and ^ 11, 3 h .
5.
Find the point that divides the
interval between ^ 3, -2 h and
^ 5, 8 h in the ratio of 1:1.
6.
Show that the line x = 3 is the
perpendicular bisector of the
interval between the points
^ -1, 2 h and ^ 7, 2 h .
7.
8.
9.
The points A ^ -1, 2 h, B ^ 1, 5 h,
C ^ 6, 5 h and D ^ 4, 2 h form a
parallelogram. Find the midpoints
of the diagonals AC and BD. What
property of a parallelogram does
this show?
The points A ^ 3, 5 h, B ^ 9, -3 h,
C ^ 5, -6 h and D ^ -1, 2 h form a
quadrilateral. Prove that the
diagonals are equal and bisect
one another. What type of
quadrilateral is ABCD?
A circle with centre ^ -2, 5 h has
one end of a diameter at ^ 4, -3 h .
Find the coordinates of the other
end of the diameter.
10. A triangle has vertices at
A ^ -1, 3 h, B ^ 0, 4 h and C ^ 2, -2 h .
(a) Find the midpoints X, Y
and Z of sides AB, AC and BC
respectively.
1
(b) Show that XY = BC,
2
1
1
XZ = AC and YZ = AB.
2
2
11. Point P ^ x, y h moves so that
the midpoint between P and
the origin is always a point on
the circle x 2 + y 2 = 1. Find the
equation of the locus of P.
12. Find the equation of the locus
of the point P ^ x, y h that is the
midpoint between all points on
the circle x 2 + y 2 = 4 and the
origin.
Gradient
The gradient of a straight line measures its slope. The gradient compares the
vertical rise with the horizontal run.
Chapter 7 Linear Functions
399
rise
Gradient = run
On the number plane, this is a measure of the rate of change of y with
respect to x.
The rate of change of y with respect to x is a very important measure
of their relationship. In later chapters you will use the gradient for many
purposes, including sketching curves, finding the velocity and acceleration
of objects, and finding maximum and minimum values of formulae.
EXAMPLES
Find the gradient of each interval.
1.
You will study the
gradient at different
points on a curve in
the next chapter.
Solution
rise
Gradient = run
2
=
3
CONTINUED
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Maths In Focus Mathematics Extension 1 Preliminary Course
2.
Solution
In this case, x is - 3 (the run is measured towards the left).
rise
Gradient = run
2
=
-3
2
=3
Positive gradient leans to the right.
Negative gradient leans to the left.
Gradient given 2 points
The gradient of the line between _ x 1, y 1 i and _ x 2, y 2 i is given by
y2 - y1
m= x -x
2
1
Proof
Chapter 7 Linear Functions
401
BC = y 2 - y 1 and AC = x 2 - x 1
rise
Gradient = run
y2 - y1
= x -x
2
1
This formula could also be
y1 - y2
written m =
x1 - x2
EXAMPLES
1. Find the gradient of the line between points ^ 2, 3 h and ^ -3, 4 h .
Solution
y2 - y1
Gradient: m = x - x
2
1
4-3
=
-3 - 2
1
=
-5
1
=5
2. Prove that points ^ 2, 3 h, ^ -2, -5 h and ^ 0, -1 h are collinear.
Solution
To prove points are collinear, we show that they have the same gradient
(slope).
CONTINUED
Collinear points lie on the
same line, so they have
the same gradients.
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Gradient of the interval between ^ -2, -5 h and ^ 0, -1 h :
y2 - y1
m= x -x
2
1
-1 - ] -5 g
=
0 - ] -2 g
-1 + 5
=
2
4
=
2
=2
Gradient of the interval between ^ 0, -1 h and ^ 2, 3 h :
y2 - y1
m= x -x
2
1
3 - ] -1 g
=
2-0
3+1
=
2
4
=
2
=2
Since the gradient of both intervals is the same, the points are collinear.
Gradient given the angle at the x-axis
The gradient of a straight line is given by
m = tan i
where i is the angle the line makes with the x-axis in the positive direction
Proof
rise
m = run
opposite
=
adjacent
= tan i
Chapter 7 Linear Functions
For an acute angle tan i 2 0.
403
For an obtuse angle tan i 1 0.
Class Discussion
1. Which angles give a positive gradient?
2. Which angles give a negative gradient? Why?
3. What is the gradient of a horizontal line? What angle does it make
with the x-axis?
4. What angle does a vertical line make with the x-axis? Can you find
its gradient?
EXAMPLES
1. Find the gradient of the line that makes an angle of 135c with the
x-axis in the positive direction.
Solution
m = tan i
= tan 135c
= -1
2. Find the angle, in degrees and minutes, that a straight line makes
with the x-axis in the positive direction if its gradient is 0.5.
Solution
m = tan i
` tan i = 0.5
i = 26c34l
Can you see why the
gradient is negative?
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7.3 Exercises
1.
2.
3.
Find the gradient of the line
between
(a) ^ 3, 2 h and ^ 1, -2 h
(b) ^ 0, 2 h and ^ 3, 6 h
(c) ^ -2, 3 h and ^ 4, -5 h
(d) ^ 2, -5 h and ^ -3, 7 h
(e) ^ 2, 3 h and ^ -1, 1 h
(f) ^ - 5, 1 h and ^ 3, 0 h
(g) ^ -2, -3 h and ^ -4, 6 h
(h) ^ -1, 3 h and ^ -7, 7 h
(i) ^ 1, -4 h and ^ 5, 5 h
(j) ^ 0, 4 h and ^ 3, -2 h
If the gradient of _ 8, y 1 i and
^ -1, 3 h is 2, find the value of y 1 .
The gradient of ^ 2, -1 h and ^ x, 0 h
is –5. Find the value of x.
4.
The gradient of a line is –1 and
the line passes through the points
^ 4, 2 h and ^ x, -3 h . Find the value
of x.
5. (a) Show that the gradient of
the line through ^ -2, 1 h and
^ 3, 4 h is equal to the gradient
of the line between the points
^ 2, -1 h and ^ 7,2 h .
(b) Draw the two lines on the
number plane. What can you say
about the lines?
6.
7.
8.
Show that the points
A ^ -1, 2 h, B ^ 1, 5 h, C ^ 6, 5 h and
D ^ 4, 2 h form a parallelogram.
Find the gradients of all sides.
The points
A ^ 3, 5 h, B ^ 9, -3 h, C ^ 5, -6 h and
D ^ -1, 2 h form a rectangle. Find
the gradients of all the sides and
the diagonals.
Find the gradients of the
diagonals of the square with
vertices A ^ -2, 1 h, B ^ 3, 1 h,
C ^ 3, 6 h and D ^ -2, 6 h .
9.
A triangle has vertices
A ^ 3, 1 h, B ^ -1, -4 h and C ^ -11, 4 h .
(a) By finding the lengths of all
sides, prove that it is a rightangled triangle.
(b) Find the gradients of sides
AB and BC.
10. (a) Find the midpoints F and
G of sides AB and AC where
ABC is a triangle with vertices
A ^ 0, 3 h, B ^ 2, -7 h and C ^ 8, -2 h .
(b) Find the gradients of FG
and BC.
11. The gradient of the line between
a moving point P ^ x, y h and the
point A ^ 5, 3 h is equal to the
gradient of line PB where B has
coordinates ^ 2, -1 h . Find the
equation of the locus of P.
12. Prove that the points ^ 3, -1 h, ^ 5, 5 h
and ^ 2, -4 h are collinear.
13. Find the gradient of the straight
line that makes an angle of 45c
with the x-axis in the positive
direction.
14. Find the gradient, to 2 significant
figures, of the straight line that
makes an angle of 42c51l with
the x-axis.
15. Find the gradient of the line that
makes an angle of 87c14l with
the x-axis, to 2 significant figures.
16. Find the angle, in degrees and
minutes, that a line with gradient
1.2 makes with the x-axis.
17. What angle, in degrees and
minutes does the line with
gradient –3 make with the x-axis
in the positive direction?
Chapter 7 Linear Functions
18. Find the exact gradient of the
line that makes an angle with the
x-axis in the positive direction of
(a) 60c
(b) 30c
(c) 120c.
19. Show that the line passing
through ^ 4, -2 h and ^ 7, -5 h
makes an angle of 135c with
the x-axis in the positive
direction.
20. Find the exact value of x with
rational denominator if the line
passing through ^ x, 3 h and ^ 2, 1 h
makes an angle of 60c with the
x-axis.
Gradient given an equation
In Chapter 5 you explored and graphed linear functions. You may have
noticed a relationship between the graph and the gradient and y-intercept of a
straight line.
Investigation
1. (i) Draw the graph of each linear function.
(ii) By selecting two points on the line, find its gradient.
(a) y = x
(b) y = 2x
(c) y = 3x
(d) y = - x
(e) y = - 2x
Can you find a pattern for the gradient of each line? Can you predict
what the gradient of y = 5x and y = - 9x would be?
2. (i) Draw the graph of each linear function.
(ii) Find the y-intercept.
(a) y = x
(b) y = x + 1
(c) y = x + 2
(d) y = x - 2
(e) y = x - 3
Can you find a pattern for the y-intercept of each line? Can you predict
what the y-intercept of y = x + 11 and y = x - 6 would be?
y = mx + b has
m = gradient
b = y-intercept
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EXAMPLES
1. Find the gradient and y-intercept of the linear function y = 7x - 5.
Solution
The equation is in the form y = mx + b where m = 7 and b = - 5.
Gradient = 7
y-intercept = - 5
2. Find the gradient of the straight line with equation 2x + 3y - 6 = 0.
Solution
First, we change the equation into the form y = mx + b.
2x + 3y - 6 = 0
2x + 3y - 6 + 6 = 0 + 6
2x + 3y = 6
2x - 2x + 3y = 6 - 2x
3y = 6 - 2x
= - 2x + 6
3y
- 2x + 6
=
3
3
- 2x 6
y=
+
3
3
2
= - x +2
3
2
m=3
2
So the gradient is - .
3
There is a general formula for finding the gradient of a straight line.
The gradient of the line ax + by + c = 0 is given by
m=-
Proof
ax + by + c = 0
by = - ax - c
ax c
y=b
b
a
`
m=b
a
b
Chapter 7 Linear Functions
EXAMPLE
Find the gradient of 3x - y = 2.
Solution
3x - y = 2
3x - y - 2 = 0
a = 3, b = - 1
a
m=b
3
=-1
=3
` gradient is 3
7.4 Exercises
1.
Find
(i) the gradient and
(ii) the y-intercept of each linear
function.
(a) y = 3x + 5
(b) f ] x g = 2x + 1
(c) y = 6x - 7
(d) y = - x
(e) y = - 4x + 3
(f) y = x - 2
(g) f ] x g = 6 - 2x
(h) y = 1 - x
(i) y = 9x
(j) y = 5x - 2
2.
Find
(i) the gradient and
(ii) the y-intercept of each linear
function.
(a) 2x + y - 3 = 0
(b) 5x + y + 6 = 0
(c) 6x - y - 1 = 0
(d) x - y + 4 = 0
(e) 4x + 2y - 1 = 0
(f) 6x - 2y + 3 = 0
(g) x + 3y + 6 = 0
(h) 4x + 5y - 10 = 0
(i) 7x - 2y - 1 = 0
(j) 5x - 3y + 2 = 0
3.
Find the gradient of the straight
line.
(a) y = 4x
(b) y = - 2x - 1
(c) y = 2
(d) 2x + y - 5 = 0
(e) x + y + 1 = 0
(f) 3x + y = 8
(g) 2x - y + 5 = 0
(h) x + 4y - 12 = 0
(i) 3x - 2y + 4 = 0
(j) 5x - 4y = 15
2
(k) y = x + 3
3
x
(l) y =
2
x
(m) y = - 1
5
2x
(n) y =
+5
7
3x
-2
(o) y = 5
x 1
(p) 2y = - +
7 3
y
(q) 3x - = 8
5
x y
(r)
+ =1
2 3
2x
(s)
- 4y - 3 = 0
3
x 2y
+
+7=0
(t)
4
3
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Equation of a Straight Line
There are several different ways to write the equation of a straight line.
General form
ax + by + c = 0
Gradient form
y = mx + b
where m = gradient and b = y-intercept
Intercept form
x y
a+b =1
where a and b are the x-intercept and y-intercept respectively
Proof
b
m = - a, b = b
`
b
y = -ax + b
y
`
b
y
x
= -a + 1
x
a+b =1
Point-gradient formula
There are two formulae for finding the equation of a straight line. One of these
uses a point and the gradient of the line.
The equation of a straight line is given by
y - y1 = m _ x - x1 i
This is a very useful
formula as it is used in
many topics in this course.
where _ x 1, y 1 i lies on the line with gradient m
Chapter 7 Linear Functions
409
Proof
Given point _ x 1, y 1 i on the line with gradient m
Let P = ^ x, y h
Then line AP has gradient
y2 - y1
m= x -x
2
1
y - y1
`
m= x-x
1
m _ x - x1 i = y - y1
Two-point formula
The equation of a straight line is given by
y - y1
y2 - y1
=
x - x1
x2 - x1
This formula is
optional as you can
use the point–gradient
formula for any
question.
where _ x 1, y 1 i and _ x 2, y 2 i are points on the line
Proof
The gradient is the
same anywhere along
a straight line.
Let P = ^ x, y h
D APQ <; D ABR
PQ
BR
So
=
AR
AQ
y - y1
y2 - y1
i.e. x - x = x - x
1
2
1
The two-point formula is not essential. The right-hand side of it is the gradient
of the line. Replacing this by m gives the point–gradient formula.
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Maths In Focus Mathematics Extension 1 Preliminary Course
EXAMPLES
1. Find the equation of the straight line with gradient -4 and passing
through the point ^ -2, 3 h .
Solution
m = -4, x 1 = -2 and y 1 = 3
Equation: y - y 1 = m (x - x 1)
y - 3 = - 4 [x - (-2)]
= - 4 (x + 2)
= - 4x - 8
`
y = - 4x - 5
or 4x + y + 5 = 0
(gradient form)
(general form)
2. Find the equation of the straight line that passes through the points
^ 2, -3 h and ^ -4, -7 h .
Solution
By two-point formula:
y - y1
y2 - y1
=
x - x1
x2 - x1
y - ] -7 g
-3 - ] -7 g
=
x - ] -4 g
2 - ] -4 g
y+7
-3 + 7
=
x+4
2+4
y+7
2
=
x+4
3
3 ^ y + 7 h = 2 ]x + 4 g
3y + 21 = 2x + 8
-2x + 3y + 13 = 0
or 2x - 3y - 13 = 0
By point-gradient method:
y2 - y1
m= x -x
2
1
-3 - ] -7 g
=
2 - ] -4 g
-3 + 7
=
2+4
2
=
3
Use one of the points, say ^ -4, -7 h .
2
m = , x 1 = -4 and y 1 = -7
3
Equation:
y - y 1 = m ( x - x 1)
y - (-7) =
2
6 x - ( - 4) @
3
Chapter 7 Linear Functions
2
( x + 4)
3
= 2 ]x + 4 g
= 2x + 8
=0
=0
y+7=
3^ y + 7h
3y + 21
` -2x + 3y + 13
or 2x - 3y - 13
3. Find the equation of the line with x-intercept 3 and y-intercept 2.
Solution
x y
Intercept form is a + = 1, where a and b are the x-intercept and
b
y-intercept respectively.
x y
`
+ =1
3 2
2x + 3y = 6
` 2x + 3y - 6 = 0
Again, the point-gradient formula can be used. The x-intercept and
y-intercept are the points ^ 3, 0 h and ^ 0, 2 h .
7.5 Exercises
1.
Find the equation of the straight
line
(a) with gradient 4 and
y-intercept -1
(b) with gradient -3 and passing
through ^ 0, 4 h
(c) passing through the origin
with gradient 5
(d) with gradient 4 and
x-intercept -5
(e) with x-intercept 1 and
y-intercept 3
(f) with x-intercept 3,
y-intercept -4
(g) with y-intercept -1 and
making an angle of 45c with the
x-axis in the positive direction
(h) with y-intercept 5 and making
an angle of 45c with the x-axis in
the positive direction.
2.
Find the equation of the straight
line that makes an angle of
135c with the x-axis and passes
through the point ^ 2, 6 h .
3.
Find the equation of the straight
line passing through
(a) ^ 2, 5 h and ^ -1, 1 h
(b) ^ 0, 1 h and ^ -4, -2 h
(c) ^ - 2, 1 h and ^ 3, 5 h
(d) ^ 3, 4 h and ^ -1, 7 h
(e) ^ -4, -1 h and ^ - 2, 0 h .
4.
What is the equation of the line
with x-intercept 2 and passing
through ^ 3, -4 h ?
5.
Find the equation of the line
(a) parallel to the x-axis and
passing through ^ 2, 3 h
(b) parallel to the y-axis and
passing through ^ -1, 2 h .
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Maths In Focus Mathematics Extension 1 Preliminary Course
6.
A straight line passing through
the origin has a gradient of - 2.
Find its equation.
7.
A straight line has x-intercept 4
and passes through ^ 0, -3 h . Find
its equation.
8.
Find the equation of the straight
line with gradient -2 that passes
through the midpoint of ^ 5, -2 h
and ^ -3, 4 h .
9.
What is the equation of the
straight line through the point
^ -4, 5 h and the midpoint of ^ 1, 2 h
and ^ -9, 4 h ?
10. What is the equation of the
straight line through the
midpoint of ^ 0, 1 h and ^ -6, 5 h
and the midpoint of ^ 2, 3 h and
^ 8, -3 h ?
Parallel and Perpendicular Lines
Parallel lines
Class Investigation
Sketch the following straight lines on the same number plane.
1. y = 2x
2. y = 2x + 1
3. y = 2x - 3
4. y = 2x + 5
What do you notice about these lines?
If two lines are parallel, then they have the same gradient. That is,
m1 = m2
Two lines that are parallel have equations
ax + by + c 1 = 0 and ax + by + c 2 = 0
Chapter 7 Linear Functions
413
Proof
a
b
a
ax + by + c 2 = 0 has gradient m 2 = b
Since m 1 = m 2, the two lines are parallel.
ax + by + c 1 = 0 has gradient m 1 = -
EXAMPLES
1. Prove that the straight lines 5x - 2y - 1 = 0 and 5x - 2y + 7 = 0 are
parallel.
Solution
5x - 2y - 1 = 0
5x - 1 = 2y
5
1
x- =y
2
2
5
`
m1 =
2
5x - 2y + 7 = 0
5x + 7 = 2 y
5
7
x+ =y
2
2
5
`
m2 =
2
5
m1 = m2 =
2
` the lines are parallel.
2. Find the equation of a straight line parallel to the line 2x - y - 3 = 0
and passing through ^ 1, -5 h .
Solution
2x - y - 3 = 0
2x - 3 = y
`
m1 = 2
For parallel lines m 1 = m 2
` m2 = 2
Equation:
y - y 1 = m (x - x 1)
y - (-5) = 2 (x - 1)
y + 5 = 2x - 2
0 = 2x - y - 7
Notice that the equations
are both in the form
5x - 2y + k = 0.
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Maths In Focus Mathematics Extension 1 Preliminary Course
DID YOU KNOW?
Parallel lines are usually thought of as lines that never meet. However, there is a whole branch
of geometry based on the theory that parallel lines meet at infinity. This is called affine
geometry. In this geometry there are no perpendicular lines.
Perpendicular lines
Class Investigation
Sketch the following pairs of straight lines on the same number plane.
1. (a) 3x - 4y + 12 = 0
2. (a) 2x + y + 4 = 0
(b) 4x + 3y - 8 = 0
(b) x - 2y + 2 = 0
What do you notice about these pairs of lines?
Gradients of perpendicular lines
are negative reciprocals of each
other.
If two lines with gradients m 1 and m 2 respectively are perpendicular, then
m 1 m 2 = -1
1
i.e. m 2 = - m
1
Proof
Let line AB have gradient m 1 = tan a .
Let line CD have gradient m 2 = tan b.
EB
EC
+CBE = 180c - a
EC
tan ] 180c - a g =
EB
EB
` cot ] 180c - a g =
EC
tan b =
^ straight angle h
Chapter 7 Linear Functions
`
So
or
tan b = cot ] 180c - a g
= - cot a
1
=tan a
1
m2 = - m
1
m 1 m 2 = -1
Perpendicular lines have equations in the form
ax + by + c 1 = 0 and bx - ay + c 2 = 0
Proof
a
b
b
bx - ay + c 2 = 0 has gradient m 2 = - - a
b
=a
a
b
m1 m2 = - # a
b
= -1
ax + by + c 1 = 0 has gradient m 1 = -
Since m 1 m 2 = -1, the two lines are perpendicular.
EXAMPLES
1. Show that the lines 3x + y - 11 = 0 and x - 3y + 1 = 0 are
perpendicular.
Solution
3x + y - 11 = 0
y = -3x + 11
m 1 = -3
`
x - 3y + 1 = 0
x + 1 = 3y
1
1
x+ =y
3
3
1
`
m2 =
3
1
m 1 m 2 = - 3#
3
= -1
Notice that the equations
are in the form
3x + y + c 1 = 0 and
x - 3y + c 2 = 0.
` the lines are perpendicular.
CONTINUED
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Maths In Focus Mathematics Extension 1 Preliminary Course
2. Find the equation of the straight line through ^ 2, 3 h perpendicular to
the line that passes through ^ -1, 7 h and ^ 3, 3 h .
Solution
Line through ^ -1, 7 h and ^ 3, 3 h:
y2 - y1
m= x -x
2
1
7-3
m1 =
-1 - 3
4
=
-4
= -1
For perpendicular lines, m 1 m 2 = - 1
i.e.
-1m 2 = - 1
m2 = 1
Equation through ^ 2, 3 h:
y - y 1 = m (x - x 1)
y - 3 = 1 (x - 2 )
=x-2
0=x-y+1
7.6 Exercises
1.
Find the gradient of the straight
line
(a) parallel to the line
3x + y - 4 = 0
(b) perpendicular to the line
3x + y - 4 = 0
(c) parallel to the line joining
^ 3, 5 h and ^ -1, 2 h
(d) perpendicular to the line with
x-intercept 3 and y-intercept 2
(e) perpendicular to the line
making an angle of 135c with the
x-axis in the positive direction
(f) perpendicular to the line
6x - 5y - 4 = 0
(g) parallel to the line making an
angle of 30c with the x-axis
(h) parallel to the line
x - 3y - 7 = 0
(i) perpendicular to the line
making an angle of 120c with the
x-axis in the positive direction
(j) perpendicular to the line
passing through ^ 4, -2 h and ^ 3, 3 h .
2.
Find the equation of each straight
line
(a) passing through ^ 2, 3 h and
parallel to the line y = x + 6
(b) through ^ -1, 5 h and parallel
to the line x - 3y - 7 = 0
(c) with x-intercept 5 and parallel
to the line y = 4 - x
(d) through ^ 3, -4 h and
perpendicular to the line y = 2x
(e) through ^ -2, 1 h and
perpendicular to the line
2x + y + 3 = 0
Chapter 7 Linear Functions
(f) through ^ 7, -2 h and
perpendicular to the line
3x - y - 5 = 0
(g) through ^ -3, -1 h and
perpendicular to the line
4x - 3y + 2 = 0 .
3.
Show that the straight lines
y = 3x - 2 and 6x - 2y - 9 = 0
are parallel.
4.
Show that lines x + 5y = 0 and
y = 5x + 3 are perpendicular.
5.
Show that lines 6x - 5y + 1 = 0
and 6x - 5y - 3 = 0 are parallel.
6.
Show that lines 7x + 3y + 2 = 0
and 3x - 7y = 0 are
perpendicular.
7.
If the lines 3x - 2y + 5 = 0 and
y = kx - 1 are perpendicular, find
the value of k.
8.
9.
Show that the line joining ^ 3, -1 h
and ^ 2, -5 h is parallel to the line
8x - 2y - 3 = 0.
Show that the points A ^ -3, -2 h,
B ^ -1, 4 h, C ^ 7, -1 h, and
D ^ 5, -7 h are the vertices of a
parallelogram.
10. The points A ^ -2, 0 h, B ^ 1, 4 h,
C ^ 6, 4 h and D ^ 3, 0 h form
a rhombus. Show that the
diagonals are perpendicular.
417
11. Find the equation of the straight
line
(a) passing through the
origin and parallel to the line
x+y+3=0
(b) through ^ 3, 7 h and parallel to
the line 5x - y - 2 = 0
(c) through ^ 0, - 2 h and
perpendicular to the line
x - 2y = 9
(d) perpendicular to the line
3x + 2y - 1 = 0 and passing
through the point ^ -2, 4 h .
12. Find the equation of the straight
line passing through ^ 6, -3 h
that is perpendicular to the line
joining ^ 2, -1 h and ^ -5, -7 h .
13. Find the equation of the line
through ^ 2, 1 h that is parallel
to the line that makes an angle
of 135c with the x-axis in the
positive direction.
14. Find the equation of the
perpendicular bisector of the
line passing through ^ 6, -3 h and
^ -2, 1 h .
15. Find the equation of the
straight line parallel to the line
2x - 3y - 1 = 0 and through the
midpoint of ^ 1, 3 h and ^ -1, 9 h .
Intersection of Lines
Two straight lines intersect at a single point ^ x, y h . The point satisfies the
equations of both lines. We find this point by solving simultaneous equations.
You may need to revise
simultaneous equations
from Chapter 3.
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Maths In Focus Mathematics Extension 1 Preliminary Course
Concurrent lines meet at a single point. To show that lines are
concurrent, solve two simultaneous equations to find the point of intersection.
Then substitute this point of intersection into the third and subsequent lines
to show that these lines also pass through the point.
EXAMPLES
1. Find the point of intersection between lines 2x - 3y - 3 = 0 and
5x - 2y - 13 = 0.
Solution
Solve simultaneous equations:
2x - 3y - 3 = 0
5x - 2y - 13 = 0
4x - 6y - 6 = 0
^ 1 h # 2:
15x - 6y - 39 = 0
^ 2 h # 3:
+ 33 = 0
^ 3 h - ^ 4 h: -11x
33 = 11x
3=x
^1h
^2h
^3h
^4h
Substitute x = 3 into ^ 1 h:
You could use a
computer spreadsheet to
solve these simultaneous
equations.
2 ^ 3 h - 3y - 3 = 0
- 3y + 3 = 0
3 = 3y
1=y
So the point of intersection is ^ 3, 1 h .
2. Show that the lines 3x - y + 1 = 0, x + 2y + 12 = 0 and
4x - 3y - 7 = 0 are concurrent.
Solution
Solve any two simultaneous equations:
3x - y + 1 = 0
x + 2y + 12 = 0
4x - 3y - 7 = 0
6x - 2y + 2 = 0
^ 1 h # 2:
2
+
4
:
7
x
+ 14 = 0
^ h ^ h
^1h
^2h
^3h
^4h
Chapter 7 Linear Functions
7x = -14
x = -2
Substitute x = -2 into ^ 1 h:
3 ^ -2 h - y + 1 = 0
-y - 5 = 0
-5 = y
So the point of intersection of (1) and (2) is ^ -2, -5 h .
Substitute ^ -2, -5 h into (3): 4x - 3y - 7 = 0
LHS = 4 ^ -2 h - 3 ^ - 5 h - 7
= -8 + 15 - 7
=0
= RHS
So the point lies on line (3)
` all three lines are concurrent.
Equation of a line through the intersection of 2 other lines
To find the equation of a line through the intersection of 2 other lines, find
the point of intersection, then use it with the other information to find the
equation.
Another method uses a formula to find the equation.
If a 1 x + b 1 y + c 1 = 0 and a 2 x + b 2 y + c 2 = 0 are 2 given lines then the
equation of a line through their intersection is given by the formula
(a 1 x + b 1 y + c 1) + k (a 2 x + b 2 y + c 2) = 0 where k is a constant
Proof
Let l 1 have equation a 1 x + b 1 y + c 1 = 0.
Let l 2 have equation a 2 x + b 2 y + c 2 = 0.
Let the point of intersection of l 1 and l 2 be P ^ x 1, y 1 h .
Then P satisfies l 1
i.e. a 1 x 1 + b 1 y 1 + c 1 = 0
P also satisfies l2
i.e. a 2 x 1 + b 2 y 1 + c 2 = 0
Substitute P into (a 1 x + b 1 y + c 1) + k (a 2 x + b 2 y + c 2) = 0
(a 1 x 1 + b 1 y 1 + c 1) + k (a 2 x 1 + b 2 y 1 + c 2) = 0
0 + k ^0h = 0
0=0
` if point P satisfies both equations l 1 and l 2 then it satisfies l 1 + kl 2 = 0.
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Maths In Focus Mathematics Extension 1 Preliminary Course
EXAMPLE
Find the equation of the line through ^ -1, 2 h that passes through the
intersection of lines 2x + y - 5 = 0 and x - 3y + 1 = 0.
Solution
Using the formula:
a 1 = 2, b 1 = 1, c 1 = -5
a 2 = 1, b 2 = -3, c 2 = 1
^ a1 x + b1 y + c1 h + k ^ a2 x + b2 y + c2 h = 0
^ 2x + y - 5 h + k ^ x - 3y + 1 h = 0
Since this line passes through ^ -1, 2 h, substitute the point into the
equation:
^ -2 + 2 - 5 h + k ^ -1 - 6 + 1 h = 0
-5 - 6k = 0
-5 = 6k
5
- =k
6
So the equation becomes:
5
^ 2x + y - 5 h - ^ x - 3y + 1 h = 0
6
6 ^ 2x + y - 5 h - 5 ^ x - 3 y + 1 h = 0
12x + 6y - 30 - 5x + 15y - 5 = 0
7x + 21y - 35 = 0
x + 3y - 5 = 0
Another way to do this example is to find the point of intersection, then
use both points to find the equation.
Substitute the value
of k back into the
equation.
7.7 Exercises
1.
Find the point of intersection of
straight lines
(a) 3x + 4y + 10 = 0 and
2x - 3y - 16 = 0
(b) 5x + 2y + 11 = 0 and
3x + y + 6 = 0
(c) 7x - 3y = 16 and
5x - 2y = 12
(d) 2x - 3y = 6 and 4x - 5y = 10
(e) x - 3y - 8 = 0 and
4x + 7y - 13 = 0
(f) y = 5x + 6 and y = - 4x - 3
(g) y = 2x + 1 and
5x - 3y + 6 = 0
(h) 3x + 7y = 12 and
4x - y - 1 6 = 0
(i) 3x - 5y = - 7 and
2x - 3y = 4
(j) 8x - 7y - 3 = 0 and
5x - 2y - 1 = 0
2.
Show that the lines
x - 2y - 11 = 0 and
2x - y - 10 = 0 intersect at the
point ^ 3, -4 h .
3.
A triangle is formed by 3
straight lines with equations
2x - y + 1 = 0, 2x + y - 9 = 0
Chapter 7 Linear Functions
and 2x - 5y - 3 = 0. Find the
coordinates of its vertices.
4.
Show that the lines
x - 5y - 17 = 0,
3x - 2y - 12 = 0 and
5x + y - 7 = 0 are concurrent.
5.
Show that the lines
x + 4y + 5 = 0, 3x - 7y + 15 = 0,
2x - y + 10 = 0 and
6x + 5y + 30 = 0 are concurrent.
6.
Find the equation of the straight
line through the origin that
passes through the intersection of
the lines 5x - 2y + 14 = 0 and
3x + 4y - 7 = 0 .
7.
Find the equation of the straight
line through ^ 3, 2 h that passes
through the intersection of
the lines 5x + 2y + 1 = 0 and
3x - y + 16 = 0.
8.
Find the equation of the straight
line through ^ -4, -1 h that
passes through the intersection
of the lines 2x + y - 1 = 0 and
3x + 5y + 16 = 0.
9.
Find the equation of the straight
line through ^ -3, 4 h that passes
through the intersection of
the lines 2x + y - 3 = 0 and
3x - 2y - 8 = 0 .
10. Find the equation of the straight
line through ^ 2, -2 h that passes
through the intersection of
the lines 2x + 3y - 6 = 0 and
3x + 5y - 10 = 0.
11. Find the equation of the straight
line through ^ 3, 0 h that passes
through the intersection of
the lines x - y + 1 = 0 and
4x - y - 2 = 0 .
12. Find the equation of the straight
line through ^ -1, -2 h that
passes through the intersection
of the lines 2x + y - 6 = 0 and
3 x + 7 y - 9 = 0.
13. Find the equation of the straight
line through ^ 1, 2 h that passes
through the intersection of
the lines x + 2y + 10 = 0 and
2x - y + 5 = 0.
14. Find the equation of the straight
line through ^ -2, 0 h that passes
through the intersection of
the lines 3x + 4y - 7 = 0 and
3 x - 2 y - 1 = 0.
15. Find the equation of the straight
line through ^ 3, -2 h that passes
through the intersection of
the lines 5x + 2y - 13 = 0 and
x - 3y + 11 = 0.
16. Find the equation of the straight
line through ^ -3, -2 h that
passes through the intersection
of the lines x + y + 1 = 0 and
3x + 2y = 0 .
17. Find the equation of the straight
line through ^ 3, 1 h that passes
through the intersection of
the lines 3x - y + 4 = 0 and
2x - y + 12 = 0.
18. Find the equation of the straight
line with gradient 3 that passes
through the intersection of
the lines 2x + y - 1 = 0 and
3x + 5y + 16 = 0.
19. Find the equation of the straight
line with gradient 2 that passes
through the intersection of
the lines 5x - 2y - 3 = 0 and
7x - 3y - 4 = 0 .
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Maths In Focus Mathematics Extension 1 Preliminary Course
20. Find the equation of the
straight line parallel to the
line 3x - y - 7 = 0 that passes
through the intersection of
the lines 3x - 2y - 10 = 0 and
4x + y - 17 = 0.
21. Find the equation of the
straight line perpendicular to
the line x + 5y - 1 = 0 that
passes through the intersection
of lines 3x - 5y - 3 = 0 and
2x + 3y + 17 = 0.
Perpendicular Distance
The distance formula d = _ x 2 - x 1 i2 + _ y 2 - y 1 i2 is used to find the distance
between two points.
Perpendicular distance is used to find the distance between a point and
a line. If we look at the distance between a point and a line, there could be
many distances.
So we choose the shortest distance, which is the perpendicular distance.
The perpendicular distance from _ x 1, y 1 i to the line ax + by + c = 0 is
A distance is always
positive, so take the
absolute value.
given by d =
Proof
| ax 1 + by 1 + c |
a2 + b2
Chapter 7 Linear Functions
Let d be the perpendicular distance of _ x 1, y 1 i from the line ax + by + c = 0.
- ax 1 - c
c
c
o
C = c 0, - m
R = e x 1,
A = b- a , 0 l
b
b
c2
c2
+
a2
b2
In D ACO, AC =
c2 b2 + c2 a2
a2 b2
=
=
PR = y 1 - e
=
c a2 + b2
ab
- ax 1 - c
b
ax 1 + by 1 + c
o
b
Why?
D ACO is similar to D PRQ
`
`
To find A and C, substitute
y = 0 and x = 0 into
ax + by + c = 0.
PQ
PR
=
AO
AC
AO . PR
PQ =
AC
ax 1 + by 1 + c
c a2 + b2
c
d=a#
'
b
ab
c _ ax 1 + by 1 + c i
ab
=
#
ab
c a2 + b2
ax 1 + by 1 + c
=
a2 + b2
All points on one side of the line ax + by + c = 0 make the numerator of
this formula positive. Points on the other side make the numerator negative.
Usually we take the absolute value of d. However, if we want to know if
points are on the same side of a line or not, we look at the sign of d.
EXAMPLES
1. Find the perpendicular distance of ^ 4, - 3 h from the line 3x - 4y - 1 = 0.
Solution
x 1 = 4, y 1 = - 3, a = 3, b = - 4, c = - 1
| ax 1 + by 1 + c |
d=
a2 + b2
| 3 ] 4 g + ] - 4 g ] -3 g + ] -1 g |
=
3 2 + ] -4 g2
CONTINUED
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Maths In Focus Mathematics Extension 1 Preliminary Course
=
| 12 + 12 - 1 |
25
23
=
5
= 4 .6
So the perpendicular distance is 4.6 units.
2. Prove that the line 6x + 8y + 20 = 0 is a tangent to the circle
x 2 + y 2 = 4.
Solution
There are three possibilities for the intersection of a circle and a straight line.
The centre of the circle x 2 + y 2 = 4 is ^ 0, 0 h and its radius is 2 units.
A tangent is perpendicular to the centre of the circle. So we prove that the
perpendicular distance from the line to the point ^ 0, 0 h is 2 units (the radius).
| ax 1 + by 1 + c |
d=
a2 + b2
| 6 (0) + 8 (0) + 20 |
=
62 + 82
| 20 |
=
100
20
=
10
=2
` the line is a tangent to the circle.
3. Show that the points ^ -1, 3 h and ^ 2, 7 h lie on the same side of the line
2 x - 3 y + 4 = 0.
Chapter 7 Linear Functions
Solution
To show that points lie on the same side of a line, their perpendicular
distance must have the same sign. We use the formula without the
absolute value sign.
d=
ax 1 + by 1 + c
a2 + b2
^ - 1, 3 h :
2 ]-1 g - 3 ]3 g + 4
d=
22 + ] - 3 g 2
-2 - 9 + 4
=
4+9
-7
=
13
^ 2, 7 h :
2 ]2 g - 3 ]7 g + 4
d=
2 2 + ] -3 g 2
4 - 21 + 4
=
4+9
- 13
=
13
Since the perpendicular distance for both points has the same sign, the
points lie on the same side of the line.
7.8 Exercises
1.
Find the perpendicular distance
between
(a) ^ 1, 2 h and 3x + 4y + 2 = 0
(b) ^ - 3, 2 h and 5x + 12y + 7 = 0
(c) ^ 0, 4 h and 8x - 6y - 1 = 0
(d) ^ - 3, - 2 h and 4x - 3y - 6 = 0
(e) the origin and
12x - 5y + 8 = 0.
2.
Find, correct to 3 significant
figures, the perpendicular
distance between
(a) ^ 1, 3 h and x + 3y + 1 = 0
(b) ^ -1, 1 h and 2x + 5y + 4 = 0
(c) ^ 3, 0 h and 5x - 6y - 12 = 0
(d) ^ 5, - 3 h and 4x - y - 2 = 0
(e) ^ - 6, - 3 h and 2x - 3y + 9 = 0.
3.
Find as a surd with rational
denominator the perpendicular
distance between
(a) the origin and the line
3x - 2y + 7 = 0
(b) ^ -1, 4 h and 2x + y + 3 = 0
(c) ^ 3, -1 h and 3x + 14y + 1 = 0
(d) ^ 2, - 6 h and 5x - y - 6 = 0
(e) ^ - 4, - 1 h and
3 x - 2 y - 4 = 0.
4.
Show that the origin
is equidistant from the
lines 7x + 24y + 25 = 0,
4x + 3y - 5 = 0 and
12x + 5y - 13 = 0.
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Maths In Focus Mathematics Extension 1 Preliminary Course
5.
6.
7.
Equidistant means that
two or more objects are
the same distance away
from another object.
8.
9.
Show that points A ^ 3, - 5 h and
B ^ -1, 4 h lie on opposite sides of
2x - y + 3 = 0.
14. Find the perpendicular distance
between ^ 0, 5 h and the line
through ^ - 3, 8 h parallel to
4x - 3y - 1 = 0.
Show that the points ^ 2, - 3 h and
^ 9, 2 h lie on the same side of the
line x - 3y + 2 = 0.
15. The perpendicular distance
between the point ^ x, -1 h and
the line 3x - 4y + 7 = 0 is
8 units. Find two possible values
of x.
Show that ^ - 3, 2 h and ^ 4, 1 h lie
on opposite sides of the line
4 x - 3 y - 2 = 0.
Show that ^ 0, - 2 h is equidistant
from the lines 3x + 4y - 2 = 0
and 12x - 5y + 16 = 0.
16. The perpendicular distance
between the point ^ 3, b h and the
line 5x - 12y - 2 = 0 is 2 units.
Find the values of b.
Show that the points ^ 8, - 3 h and
^ 1, 1 h lie on the same side of the
line 6x - y + 4 = 0.
17. Find m if the perpendicular
distance between ^ m, 7 h and the
line 9x + 12y + 6 = 0 is 5 units.
10. Show that ^ - 3, 2 h and ^ 4, 1 h lie
on opposite sides of the line
2x + y - 2 = 0.
18. Prove that the line
3x - 4y + 25 = 0 is a tangent to
the circle with centre the origin
and radius 5 units.
11. Show that the point ^ 3, - 2 h
is the same distance from the
line 6x - 8y + 6 = 0 as the
point ^ - 4, -1 h is from the line
5x + 12y - 20 = 0.
19. Show that the line
3x - 4y + 12 = 0 does not cut
the circle x 2 + y 2 = 1.
20. The sides of a triangle are formed
by the lines with equations
2x - y - 7 = 0, 3x + 5y - 4 = 0
and x + 3y - 4 = 0.
(a) Find the vertices of the
triangle.
(b) Find the exact length of all
the altitudes of the triangle.
12. Find the exact perpendicular
distance with rational
denominator from the point
^ 4, 5 h to the line with
x-intercept 2 and y-intercept -1.
13. Find the perpendicular distance
from ^ - 2, 2 h to the line passing
through ^ 3, 7 h and ^ -1, 4 h .
Angle Between Two Lines
The acute angle i between two straight lines is given by
tan i =
m1 - m2
1 + m1 m2
where m 1 and m 2 are the gradients of the lines
Chapter 7 Linear Functions
Proof
Let line l 1 have gradient m 1 and line l 2 have gradient m 2 .
Then m 1 = tan b and m 2 = tan a
b = a + i ^ exterior angle of DABC h
` i=b-a
tan i = tan (b - a )
tan b - tan a
=
1 + tan b tan a
m1 - m2
=
1 + m1 m2
When tan i is positive, i is acute.
When tan i is negative, i is obtuse.
` for the acute angle between lines l 1 and l 2,
tan i =
Note: the denominator
cannot be zero, so
m 1 m 2 ! -1. So this
formula doesn’t work for
perpendicular lines.
m1 - m2
1 + m1 m2
EXAMPLES
1. Find the acute angle between the lines 3x - 2y + 1 = 0 and
x - 3 y = 0.
Solution
3x - 2y + 1 = 0
3x + 1 = 2y
3
1
x+ =y
2
2
3
m1 =
So
2
x - 3y = 0
x = 3y
1
x=y
3
1
m2 =
So
3
m1 - m2
tan i =
1 + m1 m2
CONTINUED
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Maths In Focus Mathematics Extension 1 Preliminary Course
3 1
2 3
=
3
1
1+ #
2
3
7
=
9
7
i = tan -1 c m
9
= 37c 52l
2. Find the obtuse angle between the lines 5x - 2y + 6 = 0 and
2x + y - 4 = 0.
Solution
9
Notice that tan -1 d - n
8
gives - 48c 22l so we need
to find the obtuse angle by
subtracting the acute angle
from 180c.
5x - 2y + 6 = 0
5x + 6 = 2y
5
x+3=y
2
5
So m 1 =
2
2x + y - 4 = 0
y = - 2x + 4
So m 2 = - 2
m1 - m2
tan i =
1 + m1 m2
5 ]
- -2g
2
=
5
1 + # ]-2 g
2
9
= 8
9
=
8
9
i = tan - 1 c m
8
= 48° 22l
This gives the acute angle.
Obtuse angle = 180c - 48c 22l
= 131c 38l
3. If the angle between the lines 2x - y - 7 = 0 and y = mx + 3 is 25c,
find two possible values of m, correct to 1 decimal place.
Solution
2x - y - 7 = 0
2x - 7 = y
`
m1 = 2
( 1)
Chapter 7 Linear Functions
y = mx + 3
`
m2 = m
m1 - m2
tan i =
1 + m1 m2
2-m
tan 25° =
1 + 2m
(2 )
There are two possibilities:
(1)
(2)
2-m
1 + 2m
tan 25c (1 + 2m) = 2 - m
tan 25c + 2m tan 25c = 2 - m
2m tan 25c + m = 2 - tan 25c
m (2 tan 25c + 1) = 2 - tan 25c
2 - tan 25c
m=
2 tan 25c + 1
Z 0.8
tan 25c =
2-m
1 + 2m
- tan 25c (1 + 2m) = 2 - m
- tan 25c - 2m tan 25c = 2 - m
- 2m tan 25c + m = 2 + tan 25c
m (- 2 tan 25c + 1) = 2 + tan 25c
2 + tan 25c
m=
- 2 tan 25c + 1
Z 36.6
- tan 25c =
7.9 Exercises
1.
Find the acute angle between the
lines
(a) 2x + y + 1 = 0 and
x+y+4=0
(b) 3x - y - 7 = 0 and
5x + y + 3 = 0
(c) x + 2y = 0 and
3x - 2y + 1 = 0
(d) x + 3y + 2 = 0 and
4x + 4y - 1 = 0
(e) 2x - 5y - 3 = 0 and
x - 5y = 0
(f) 3x + y + 1 = 0 and
4x + 7y + 2 = 0
(g) 2x - 7y - 1 = 0 and
3x + 2y - 4 = 0
(h) 2x + 2y + 1 = 0 and
x + 2y = 4
(i) 3x + 4y + 1 = 0 and
5x - 2y - 2 = 0
(j) x - 2y - 3 = 0 and
6 x - 3 y + 4 = 0.
2.
Find the obtuse angle between
the lines
(a) 4x + y + 2 = 0 and
x+y-1=0
(b) 2x - 3y - 9 = 0 and
x + 2y + 4 = 0
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Maths In Focus Mathematics Extension 1 Preliminary Course
(c) x + 6y = 2 and
2x - 4y + 3 = 0
(d) 5x + 2y + 1 = 0 and
4x + y - 7 = 0
(e) 4x - 2y - 7 = 0 and
x - 3 y = 0.
3.
Find the acute angle between the
line 2x - 5y + 1 = 0 and the line
joining ^ -1, 2 h and ^ 5, 3 h .
4.
Find the acute angle between the
line joining ^ 3, 2 h and ^ -1, 4 h
and the line joining ^ 0, 5 h and
^ 2, - 7 h .
5.
6.
A ^ 2, -1 h, B ^ - 3, 4 h and C ^ 1, - 5 h
form the vertices of a triangle.
Find the interior angles of the
triangle.
Find two possible values of m
if the lines 2x + y - 5 = 0 and
y = mx + 1 intersect at an angle
of 45c.
7.
Lines y = mx + 2 and y = 5x - 9
intersect at an acute angle whose
2
tangent is . Find the possible
5
values of m.
8.
Find the values of k if
the angle between the
lines 6x - 3y - 4 = 0 and
kx - y + 5 = 0 is 58c.
9.
A ^ 0, 0 h, B ^ 1, 2 h, C ^ 5, 2 h and
D ^ 4, 0 h form the vertices of a
parallelogram.
(a) By finding all the interior
angles, show that opposite angles
are equal.
(b) Find the obtuse angle
between the diagonals of the
parallelogram.
10. By calculating the interior angles,
show that D ABC with vertices
A ^ 7, 1 h, B ^ -1, -1 h and C ^ 5, -7 h
is an isosceles triangle.
Ratios
You have a formula for the midpoint which divides an interval in half.
Sometimes we may want to divide an interval into a ratio that is not a half.
Here is a formula that we can use to divide an interval into any internal or
external ratio.
The coordinates of a point P that divides the interval between points
_ x 1, y 1 i and _ x 2, y 2 i in the ratio m: n respectively are given by
mx 2 + nx 1
my 2 + ny 1
x=
and y =
m+n
m+n
Proof
431
Chapter 7 Linear Functions
Let P ^ x, y h be the point dividing the interval AB into the ratio m:n.
Then
m
AP
= n
PB
Draw ADC parallel to the x-axis.
Then AD = x - x 1 and DC = x 2 - x.
PD < BC
AP
AD
`
=
^ intercepts have equal ratios h
PB
DC
x - x1
m
`
n = x2 - x
m _ x2 - x i = n _ x - x1 i
mx 2 - mx = nx - nx 1
mx 2 + nx 1 = mx + nx
= x ]m + n g
mx 2 + nx 1
=x
m+n
Similarly, by drawing AEF perpendicular to the x-axis, we can show that
my 2 + ny 1
y=
.
m+n
If P divides the interval internally in the ratio m: n, then the ratio is
positive and P lies on AB.
A ratio of 1:1 gives the
midpoint
x =
x1 + x2
2
,y =
y1 + y2
2
.
If P divides the interval externally in the ratio m:n, then the ratio is
negative and P lies outside AB.
m and n are measured in
opposite directions so they
have opposite signs.
EXAMPLES
1. Divide AB into the ratio 3:4 where A is ^ 6, - 2 h and B is ^ - 7, 5 h .
Solution
CONTINUED
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Maths In Focus Mathematics Extension 1 Preliminary Course
mx 2 + nx 1
m+n
]
3 -7 g + 4 ] 6 g
=
3+4
3
=
7
my 2 + ny 1
y=
m+n
3 ] 5 g + 4 ] -2 g
=
3+4
7
=
7
=1
x=
3
` P = c ,1m
7
2. If A is ^ - 2, -1 h and B is ^ 1, 5 h, find the coordinates of the point P that
divides AB externally in the ratio 2:5.
Solution
You could use - 2: 5 instead
and would still get the same
answer for P.
Let the ratio be 2: -5.
mx 2 + nx 1
x=
m+n
2 (1) + [- 5 (- 2)]
=
2 + ( - 5)
12
=
-3
= -4
my 2 + ny 1
y=
m+n
2 (5) + [- 5 (-1)]
=
2 + ( - 5)
15
=
-3
= -5
` P = ^ - 4, - 5 h
Chapter 7 Linear Functions
7.10
1.
2.
3.
Exercises
Divide these intervals internally.
(a) ^ -1, 5 h and ^ 0, - 4 h in the
ratio 2:3
(b) ^ 3, - 2 h and ^ 2, 5 h in the
ratio 4:1
(c) ^ - 3, 3 h and ^ - 2, 1 h in the
ratio 5:4
(d) ^ 3, -1 h and ^ 7, - 2 h in the
ratio 2:5
(e) ^ - 2, 1 h and ^ 5, - 4 h in the
ratio 7:3
(f) ^ - 2, 0 h and ^ - 6, 3 h in the
ratio 3:1
(g) ^ 4, 9 h and ^ - 4, 1 h in the
ratio 1:6
(h) ^ - 3, 0 h and ^ - 5, - 6 h in the
ratio 2:9
(i) ^ 2, 5 h and ^ - 3, -1 h in the
ratio 4:3
(j) ^ 1, 1 h and ^ 3, - 7 h in the
ratio 1:2.
Divide these intervals externally.
(a) ^ - 2, 3 h and ^ 6, 1 h in the
ratio 1:5
(b) ^ 4, 0 h and ^ - 3, - 5 h in the
ratio 2:7
(c) ^ - 1, 1 h and ^ 4, 7 h in the
ratio 4:3
(d) ^ 0, - 2 h and ^ 8, 3 h in the
ratio 3:1
(e) ^ - 5, 2 h and ^ 4, 4 h in the
ratio 5:4
(f) ^ 7, -1 h and ^ 0, 1 h in the
ratio 2:9
(g) ^ - 2, 2 h and ^ 6, 7 h in the
ratio 1:3
(h) ^ 1, 3 h and ^ 7, 2 h in the ratio 4:1
(i) ^ - 4, 0 h and ^ 5, - 5 h in the
ratio 6:7
(j) ^ 2, - 3 h and ^ 7, 7 h in the
ratio 8:3.
A ^ 0, 0 h, B ^ 1, 3 h and C ^ 3, 0 h are
the vertices of a triangle.
(a) Find the coordinates of point E,
which divides AB internally in the
ratio 2:1.
(b) Find the coordinates of point F,
which divides CB internally in
the ratio 2:1.
(c) Hence prove that AC = 3EF.
4.
5.
6.
7.
8.
9.
Divide the interval AB where
A = ^ 3, 2 h and B = ^ - 1, 6 h into
three equal parts.
A has coordinates ^ - 2, 5 h and
B has coordinates ^ 4, -3 h . Find
the length of PQ if P divides AB
internally in the ratio 3:2 and
Q divides AB externally in the
ratio 3:2.
An interval AB is divided
internally at P in the ratio 5:4. If
A is ^ - 1, 2 h and P is ^ 5, - 6 h, find
the coordinates of B.
The point ^ 5, 5 h divides the
interval between ^ - 1, p h and
^ q, 6 h in the ratio 2:5. Find the
value of p and q.
A triangle is formed with vertices
A ^ 5, 6 h, B ^ 0, - 4 h and C ^ - 3, 3 h .
(a) Find the point of intersection
of its medians.
(b) If D, E and F are the
midpoints of AB, AC and BC,
divide the intervals CD, BE and
AF in the ratio 2:1. What property
of medians does this show?
If ^ 0, 0 h divides the interval AB
where A = ^ a, b h and B = ^ 4, 9 h in
the external ratio of 2:1, find the
value of a and b.
10. P divides the interval between the
point ^ 2, 3 h and the intersection
of lines 2x - 3y + 19 = 0 and
5x + 2y = 0 in the ratio of 4:5.
Find the coordinates of P.
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Maths In Focus Mathematics Extension 1 Preliminary Course
Test Yourself 7
1.
Find the distance between points ^ - 1, 2 h
and ^ 3, 7 h .
2.
What is the midpoint of the origin and
the point ^ 5, - 4 h ?
3.
Find the gradient of the straight line
(a) passing through ^ 3, -1 h and ^ - 2, 5 h
(b) with equation 2x - y + 1 = 0
(c) making an angle of 30c with the
x-axis in the positive direction
(d) perpendicular to the line
5 x + 3 y - 8 = 0.
4.
5.
6.
Find the equation of the linear function
(a) passing through ^ 2, 3 h and with
gradient 7
(b) parallel to the line 5x + y - 3 = 0
and passing through ^ 1, 1 h
(c) through the origin, and
perpendicular to the line 2x - 3y + 6 = 0
(d) through ^ 3, 1 h and ^ - 2, 4 h
(e) with x-intercept 3 and y-intercept –1.
Find the perpendicular distance between
^ 2, 5 h and the line 2x - y + 7 = 0 in surd
form with rational denominator.
Prove that the line between ^ -1, 4 h
and ^ 3, 3 h is perpendicular to the line
4x - y - 6 = 0.
7.
Find the x- and y-intercepts of
2x - 5y - 10 = 0.
8.
(a) Find the equation of the straight
line l that is perpendicular to the line
1
y = x - 3 and passes through ^ 1, -1 h .
2
(b) Find the x-intercept of l.
(c) Find the exact distance from ^ 1, -1 h
to the x-intercept of l.
9.
Prove that lines y = 5x - 7 and
10x - 2y + 1 = 0 are parallel.
10. Find the equation of the straight
line passing through the origin and
parallel to the line with equation
3x - 4y + 5 = 0.
11. Find the point of intersection between
lines y = 2x + 3 and x - 5y + 6 = 0.
12. The midpoint of ^ a, 3 h and ^ - 4, b h is
^ 1, 2 h . Find the values of a and b.
13. Find the acute angle between the lines
2x - 5y + 1 = 0 and x + y - 7 = 0 to the
nearest minute.
14. Show that the lines x - y - 4 = 0,
2x + y + 1 = 0, 5x - 3y - 14 = 0 and
3x - 2y - 9 = 0 are concurrent.
15. Divide the interval between points
^ 3, - 4 h and ^ 2, 2 h in the ratio 4:5.
16. A straight line makes an angle of 153c 29l
with the x-axis in the positive direction.
What is its gradient, to 3 significant
figures?
17. The perpendicular distance from ^ 3, - 2 h
to the line 5x - 12y + c = 0 is 2. Find
2 possible values of c.
18. Find the equation of the straight line
through ^ 1, 3 h that passes through the
intersection of the lines 2x - y + 5 = 0
and x + 2y - 5 = 0.
19. Divide the interval between ^ 0, 5 h and
^ - 2, 4 h in the external ratio of 2:3.
20. The gradient of the line through ^ 3, - 4 h
and ^ x, 2 h is −5. Evaluate x.
21. Find the obtuse angle between the lines
3x - y + 3 = 0 and 2x + 5y - 1 = 0.
Chapter 7 Linear Functions
22. Show that the points ^ - 2, 1 h and
^ 6, 3 h are on opposite sides of the line
2 x - 3 y - 1 = 0.
24. Find the equation of the line with
x-intercept 4 that makes an angle of 45c
with the x-axis.
23. Find the acute angle between the lines
y = 3x - 4 and y = 5 - x.
25. Find the equation of the line with
y-intercept - 2 and perpendicular to the
line passing through ^ 3, -2 h and ^ 0, 5 h .
Challenge Exercise 7
1.
If points ^ - 3k, 1 h, ^ k - 1, k - 3 h and
^ k - 4, k - 5 h are collinear, find the
value of k.
2.
Find the equation, in exact form, of the
line passing through _ 3 , -2 i that makes
an angle of 30c with the positive x-axis.
3.
Find the equation of the circle whose
centre is at the origin and with tangent
x - 3y + 9 = 0.
4.
ABCD is a rhombus where
A = ^ - 3, 0 h, B = ^ 0, 4 h, C = ^ 5, 4 h and
D = ^ 2, 0 h . Prove that the diagonals are
perpendicular bisectors of one another.
5.
6.
7.
8.
and the point of intersection of lines
3x - 7y = 15 and 4x - y = - 5.
9.
Find the magnitude of the angle, in
degrees and minutes, that the line
joining ^ -1, 3 h and ^ 2, - 4 h makes with
the x-axis in the positive direction.
10. Find the equation of the line that passes
through the point of intersection of lines
2x + 5y + 19 = 0 and 4x - 3y - 1 = 0
that is perpendicular to the line
3x - 2y + 1 = 0.
11. Prove A ^ 2, 5 h, B ^ - 4, 5 h and C ^ -1, 2 h are
the vertices of a right-angled isosceles
triangle.
Prove that the points _ -1, 2 2 i,
_ 3 , - 6 i and _ - 5 , 2 i all lie on a
circle with centre the origin. What are
the radius and equation of the circle?
12. Find the coordinates of the centre of
a circle that passes through points
^ 7, 2 h, ^ 2, 3 h and ^ -4, -1 h .
Find the exact distance between the
parallel lines 3x + 2y - 5 = 0 and
3x + 2y = 1.
13. If ax - y - 2 = 0 and bx - 5y + 11 = 0
intersect at the point ^ 3, 4 h, find the
values of a and b.
A straight line has x-intercept A ^ a, 0 h
and y-intercept B ^ 0, b h, where a and b are
positive integers. The gradient of line AB
is -1. Find +OBA where O is the origin
and hence prove that a = b.
14. Find the equation of the straight line
through ^ 3, -4 h that is perpendicular to
the line with x-intercept and y-intercept
−2 and 5 respectively.
Find the exact perpendicular distance
between the line 2x + 3y + 1 = 0
15. Find the acute angle between the straight
lines with equations 3x - y = 5 and
2x - 4y + 1 = 0.
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Maths In Focus Mathematics Extension 1 Preliminary Course
16. Find the exact equation of the straight
line through the midpoint of ^ 0, - 5 h,
and ^ 4, -1 h that is perpendicular to the
line that makes an angle of 30c with the
x-axis.
17. Point P ^ x, y h moves so that it is
equidistant from points A ^ 1, 4 h and
B ^ - 2, 7 h . By finding the distances AP and
BP, find the equation of the locus of P.
18. Find the value of b if the lines
2x - y + 1 = 0 and bx - 7y + 5 = 0 make
an angle of 45c at their intersection.
19. Find the coordinates of trisection of the
interval between ^ 3, -1 h and ^ 1, - 5 h .
20. Prove that if two lines with gradients
m 1 and m 2 meet at an angle of
45c, then m 1 m 2 = m 1 - m 2 - 1 or
m 1 m 2 = m 2 - m 1 - 1.
21. A and B have coordinates ^ 1, 3 h and
^ - 4, 7 h respectively. If P divides AB in the
external ratio of p:1, find the coordinates
of P in terms of p.
22. (a) Show that the point ^ - 7, 7 h lies on
the line joining A ^ - 2, 0 h and B ^ 3, - 7 h .
(b) Find the ratio in which the point
divides AB.
23. The interval AB where A = ^ - 5, 3 h and
B = ^ x, y h is divided by point P in the
ratio of 3:2. If the point P has coordinates
^ 8, - 9 h, find values for x and y.
24. The angle between straight lines
2x - 3y = 0 and mx + 4y = 9 is 32c 51l.
Find the value of m correct to
2 significant figures.
25. Given points A ^ 1, 0 h, B ^ 2, 5 h and C ^ 9, 0 h
are the vertices of a triangle,
(a) find the coordinates of P that divide
AB in the ratio 2:1
(b) find the coordinates of Q that divide
CB in the ratio 2:1
(c) prove PQ < AC
(d) find the coordinates of R that divide
AC in the ratio 2:1
(e) prove PR < BC.