Straight Lines
hsn.
Higher
Mathematics
Straight Lines
Contents
Straight Lines
1 The Distance Between Points 2 The Midpoint Formula 3 Gradients 4 Collinearity 5 Gradients of Perpendicular Lines 6 The Equation of a Straight Line 7 Medians 8 Altitudes 9 Perpendicular Bisectors 10 Intersection of Lines 11 Concurrency
1
A 1 A 3 A 4 A 6 A 7 A 8 A 11 A 12 A 13 A 14 A 17
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Higher Mathematics
Straight Lines
Straight Lines
1 The Distance Between Points
A
Points on Horizontal or Vertical Lines
It is relatively straightforward to work out the distance between two points which lie on a line parallel to the x or yaxis.
y d
( x1, y1 )
O
(x2, y2 )
x
In the diagram to the left, the points
( x1, y1 ) and ( x2, y2 ) lie on a line parallel
to the xaxis, i.e. y1 = y2 .
The distance between the points is simply the difference in the xcoordinates, i.e. d= x2  x1 where x2 > x1.
y
(x2, y2 )
d
In the diagram to the left, the points
( x1, y1 ) and ( x2, y2 ) lie on a line parallel
to the yaxis, i.e. x1 = x2 .
O
( x1, y1 )
x
The distance between the points is simply the difference in the ycoordinates, i.e.
d= y2  y1 where y2 > y1 .
EXAMPLE
1. Calculate the distance between the points (7,  3) and (16,  3) .
The distance is 16  (7)
= 16 + 7 = 23 units.
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CfE Edition
Higher Mathematics
Straight Lines
The Distance Formula
The distance formula gives us a method for working out the length of the
straight line between any two points. It is based on Pythagoras's Theorem.
y
( x1, y1 )
O
(x2, y2 )
d
y2  y1
x2  x1 x
Note The " y2 y1" and " x2 x1 " come from the method above.
The distance d between the points ( x1, y1 ) and ( x2, y2 ) is d = ( x2  x1 )2 + ( y2  y1 )2 units.
EXAMPLES
2. A is the point (2, 4) and B(3,1) . Calculate the length of the line AB.
The length is ( x2  x1 )2 + ( y2  y1 )2 = (3  (2))2 + (1  4)2
= 52 + (3)2
= 25 + 9
= 34 units.
( ) 3. Calculate the distance between the points
1 2
,

15 4
and (1, 1) .
The distance is ( x2  x1 )2 + ( y2  y1 )2
( ) ( ) =
1

1 2
2+
1
+
15 4
2
( ) ( ) =

2 2

1 2
2+

4 4
+
15 4
2
( ) ( ) =

3 2
2
+
11 2 4
=
9 4
+
121 16
Note You need to become confident working with fractions and surds ? so practise!
=
36 16
+
121 16
=
157 16
=
157 4
units.
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Page 2
CfE Edition
Higher Mathematics
Straight Lines
2 The Midpoint Formula
A
The point halfway between two points is called their midpoint. It is calculated as follows.
The midpoint of
( x1, y1 )
and
(x2, y2 )
is
x1
+ 2
x2
,
y1 + 2
y2
.
It may be helpful to think of the midpoint as the "average" of two points.
EXAMPLES
1. Calculate the midpoint of the points (1,  4) and (7, 8).
The midpoint is
x1
+ 2
x2
,
y1 + 2
y2
=
7
+ 2
1,
8
+
( 4 )
2
= (4, 2).
Note Simply writing "The midpoint is (4, 2)" would be acceptable in an exam.
2. In the diagram below, A (9,  2) lies on the circumference of the circle with centre C(17,12) , and the line AB is the diameter of the circle. Find the coordinates of B.
B
C
A
Since C is the centre of the circle and AB is the diameter, C is the midpoint of AB. Using the midpoint formula, we have:
(17, 12 )
=
9
+ 2
x
,
2 + 2
y
where B is the point ( x, y ) .
By comparing x and ycoordinates, we have:
9 + x = 17 and 2
2 + y = 12 2
9 + x =34
2 + y =24
x = 25
So B is the point (25, 26) .
y = 26.
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CfE Edition
Higher Mathematics
Straight Lines
3 Gradients
A
Consider a straight line passing through the points ( x1, y1 ) and ( x2, y2 ):
y
(x2, y2 )
Note
( x1, y1 )
x2  x1
y2  y1
" " is the Greek letter "theta". It is often used to stand for an angle.
O
x
The gradient m of the line through ( x1, y1 ) and ( x2, y2 ) is
=m
change change in
ihnovr= iezrotinctaallhdeiisgthant ce
y2  y1 x2  x1
for x1 x2 .
Also, sin= ce tan O= pposite y2  y1 we obtain: Adjacent x2  x1
m = tan where is the angle between the line and the positive direction of the xaxis.
positive direction x
Note
As a result of the above definitions: lines with positive gradients slope
up, from left to right;
lines with negative gradients slope down, from left to right;
lines parallel to the xaxis have a lines parallel to the yaxis have an
gradient of zero;
undefined gradient.
We may also use the fact that:
Two distinct lines are said to be parallel when they have the same gradient (or when both lines are vertical).
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Higher Mathematics
Straight Lines
EXAMPLES
1. Calculate the gradient of the straight line shown in the diagram below. y
O m = tan = tan32? = 0?62 (to 2 d.p.).
32? x
2. Find the angle that the line joining P(2,  2) and Q (1, 7) makes with the positive direction of the xaxis.
The line has gradien= t m
y2 =  y1 x2  x1
7= + 2 1+ 2
3.
And so m = tan
tan = 3
= tan1= (3) 71?57? (to 2 d.p.).
3. Find the size of angle shown in the diagram below. y
m=5
O
x
We need to be careful because the in the question is not the in
"m = tan ".
So we work out the angle a and use this to find
y
: a = tan1 (m)
m=5
a
= tan1 (5) = 78?690?.
O
x
So = 90?  78?690?= 11?31? (to 2 d.p.).
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CfE Edition
Higher Mathematics
Straight Lines
4 Collinearity
A
Points which lie on the same straight line are said to be collinear.
To test if three points A, B and C are collinear we can: 1. Work out mAB .
2. Work out mBC (or mAC ).
3. If the gradients from 1. and 2. are the same then A, B and C are collinear. C
B
mBC
mAB = mBC so A, B and C are collinear.
mAB A
If the gradients are different then the points are not collinear.
C
B
mBC
mAB A
mAB mBC so A, B and C are not collinear.
This test for collinearity can only be used in two dimensions.
EXAMPLES
1. Show that the points P(6, 1) , Q (0, 2) and R (8, 6) are collinear.
mPQ=
2  (1)=
0  (6)
36=
1 2
mQR=
6  2= 80
84=
1 2
Since mPQ = mQR and Q is a common point, P, Q and R are collinear.
2. The points A (1,1), B(1,k ) and C(5,7) are collinear.
Find the value of k. Since the points are collinear mAB = mAC :
k  (1) = 7  (1) 1 1 5 1
k +1 = 8 2 4 k +1 = 2 ? (2)
k = 5.
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CfE Edition
Higher Mathematics
Straight Lines
5 Gradients of Perpendicular Lines
A
Two lines at rightangles to each other are said to be perpendicular.
If perpendicular lines have gradients m and m then m ? m = 1.
Conversely, if m ? m = 1 then the lines are perpendicular.
The simple rule is: if you know the gradient of one of the lines, then the gradient of the other is calculated by inverting the gradient (i.e. "flipping" the fraction) and changing the sign. For example:
if
m=
2 3
then
m
=

3 2
.
Note that this rule cannot be used if the line is parallel to the x or yaxis.
? If a line is parallel to the xaxis (m = 0), then the perpendicular line is parallel to the yaxis ? it has an undefined gradient.
? If a line is parallel to the yaxis then the perpendicular line is parallel to the xaxis ? it has a gradient of zero.
EXAMPLES
1. Given that T is the point (1,  2) and S is (4, 5) , find the gradient of a
line perpendicular to ST.
mST
=
5  (2) 4 1
=

7 5
So
m
=
5 7
since
mST ? m
= 1 .
2. Triangle MOP has vertices M(3, 9) , O(0, 0) and P(12, 4) .
Show that the triangle is rightangled.
Sketch: M(3, 9)
O(0, 0)
mOM
=
90 3  0
mMP
=
94 3 12
mOP
=
40 12  0
P(12, 4) = 3
=

5 15
=
1 3
=

1 3
Since mOM ? mOP = 1 , OM is perpendicular to OP which means MOP is rightangled at O.
Note The converse of Pythagoras's Theorem could also be used here:
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