THE DISTANCE FORMULA
A is the point (x1 , y1) and B is the point (x2 , y2)
By Pythagoras’ Theorem:
AB2 = AC2 + CB2
AB2 = (x2 – x1)2 + (y2 – y1)
2
AB = [(x2 – x1)2 + (y2 – y1)
2]
So the DISTANCE formula is:
AB = [(x2 – x1)2 + (y2 – y1)
2]
A
By
xO
THE MIDPOINT FORMULA
A is the point (x1 , y1) and B is the point (x2 , y2)
The Midpoint, M, is given by:
( )A
By
xO
Mx1 + x2
2
y1 + y2
2,
EXAMPLE 1 a) Calculate the length of the line joining pts A(3 , -4) & B(-5 , 7).
AB = [(x2 – x1)2 + (y2 – y1)
2]
AB = [(-5 – 3)2 + (7 – (-4))2]
AB = (64 + 121)
AB = (185)AB = 13.6 units
b) Find the midpoint of the line above.
( )x1 + x2
2
y1 + y2
2M = ,
( )3 + (-5)
2M = , -4 + 7
2
M = (-1 , 1.5)
EXAMPLE 2 Prove that the triangle with verticesA(-7 , 9), B(3 , 13) and C(2 , 1)is isosceles.
AC = [(x2 – x1)2 + (y2 – y1)
2]
AC = [(2 – (-7))2 + (1 – 9)2]
AC = (81 + 64)
AC = 145
AB = [(3 – (-7))2 + (13 – 9)2]
AB = (100 + 16)
AB = 116
BC = [(2 – 3)2 + (1 – 13)2]
BC = (1 + 144)
BC = 145
SinceAC = BC,
triangle ABC is isosceles.
GRADIENT OF A STRAIGHT LINE
Gradient =Change in y
Change in x
mAB = y2 – y1
x2 – x1
Also,tan =
Opposite
Adjacent
y2 – y1
x2 – x1
=mAB
A is the point (x1 , y1) and B is the point (x2 , y2)
A
By
xO
So, mAB = tan
mAB =y2 – y1
x2 – x1
mAB = tan
where is the angle between the line and the positive direction of the x-axis.
If lines are Parallel then they have equal gradients!If lines have equal gradients then they are parallel!!
Points that lie on the same straight line are said to be COLLINEAR.
To prove COLLINEARITY show that they have a common point and that they have equal gradients, i.e. PARALLEL!
x
Line slopesUP fromLEFT to RIGHT
tan is POSITIVE
GRADIENTis
POSITIVE
Line isHORIZONTAL
= 0
tan = 0
GRADIENT is
ZERO
Line slopesDOWN from
LEFT to RIGHT
tan is NEGATIVE
GRADIENTis
NEGATIVE
Line is VERTICAL
= 90
tan NOT DEFINED
GRADIENT is NOT
DEFINED
mAB = tanmPQ =
y2 – y1
x2 – x1
8 – (-7)
6 – (–4)=
= 15
10=
= tan
= tan-1
EXAMPLE 3 Calculate the gradient of the line joining points P(-4 , -7) & Q(6 , 8) and calculate the angle the line makes with the x-axis.
3
2
3
23
2
= 56.3º
mAB =y2 – y1
x2 – x1
3 – 6
-1 – 8=
= -3
-9=
EXAMPLE 4 Show that points A(8 , 6), B(–1 , 3) and C(-4 , 2) are COLLINEAR.
1
3
mBC =y2 – y1
x2 – x1
2 – 3
-4 –(-1)=
= -1
-3= 1
3
Since mAB = mBC AB is parallel to BC.Also, B is a common point so A, B and C
are Collinear.
PERPENDICULAR LINES
O x
y
A(a , b)B(-b , a)
Line OA is rotated 90o
anti-clockwise to OB.
mOA =b - 0
a - 0
b
a=
mOB =a - 0
-b - 0
a
-b=
Notice: mOA x mOB
b
a=
a
-bx = -1
If m1 x m2 =-1,then the lines arePERPENDICULAR.
If the lines arePERPENDICULAR,then m1 x m2 =-1
m1 m2
2
¼
-¾
-½
-4
43
m1 and m2 are the RECIPROCALS of the other
(with the sign changed).
You need only remember this simple rule:
turn upside down andchange the sign.
EXAMPLE 5 Write down the gradient of the lines PERPENDICULAR to the following:
a) y = 8x + 5
m = 8
mperp = -⅛
b) y = -4x - 7
m = -4
mperp = ¼
c) 6y = 5x – 12
m = 5/6
mperp = -6/5
y = 5/6x - 2
EXAMPLE 6 Prove that the triangle with vertices at A(4 , 1), B(7 , 5) and C(0 , 4) is right-angled.
mAB =y2 – y1
x2 – x1
5 – 1
7 – 4=
=4
3
mAC =4 – 1
0 – 4
=3
-4
Since mAB x mAC =-1,then the lines arePERPENDICULAR.
So ∆ABC must beright-angled at A
EQUATION OF A STRAIGHT LINE
A is the point (0 , c) and B is the point (x , y)
A
By
xO
mAB =y2 – y1
x2 – x1
y – c
x – 0=
m = y - c
xmx = y - c
So we get:
y = mx + c
y-InterceptGradient
a) m = –5 through Pt(0 , 4) y = -5x + 4
EXAMPLE 7 Write down the equations below:
b) m = 3 through Pt(0 , -2) y = 3x - 2
c) m = -1 through Pt(0 , 0) y = -x
EXAMPLE 8 Find the equation of the line passing through A(0 , -5) & B(-3 , 1).
mAB =y2 – y1
x2 – x1
= 6-3
=1 – (-5)-3 – 0
= -2
Equation:
y = -2x - 5
EQUATION OF A STRAIGHT LINE 2
A is the point (a , b) and B is the point (x , y)
mAB =y2 – y1
x2 – x1
y – b
x – am =
m(x – a) = y - b
So we get:
y - b = m(x – a)
A
By
xO
Straight Line, Gradient m,passing through the
point (a , b)
To find the equation of a line now, we need only know the GRADIENT and ONE OTHER POINT –
we DO NOT require the y-intercept.
This version is much better than y = mx + c , because often it is difficult (and unnecessary) to write the
equation in this form.
EXAMPLE 9 Find the equation of the line passing through A(2 , 3) & B(6 , 6).
mAB =y2 – y1
x2 – x1
= 34
=6 – 36 – 2
y – b = m(x – a)
y – 3 = ¾(x – 2)
4y – 12 = 3(x - 2)
4y – 12 = 3x – 6
4y = 3x + 6
EXAMPLE 10
Find the equation of the straight line parallel to the line 2x + y + 3 = 0, and which passes through the point (3 , 4).
y – b = m(x – a)
y – 4 = -2(x – 3)
y – 4 = -2x + 6
y = -2x + 10
2x + y + 3 = 0
y = -2x – 3
So m =-2
MEDIANS AND ALTITUDES
MEDIAN A straight line from one vertex (corner) of a triangle to the
MIDPOINT of the opposite side.e.g. The line AP is a Median since
P is the midpoint of BC.
ALTITUDE A straight line from one vertex of a
triangle PERPENDICULAR to the opposite side.e.g. The line CQ is an Altitude as it is perpendicular to AB.
O
y
x
A
C
B
Q
P
EXAMPLE 11 Points P(4 , -4), Q(11 , 10) &R(-9 , 6) form a ∆. The median PS & altitude RT cross at K. Find K.P
R
Q
S
( )x1 + x2
2
y1 + y2
2S = ,
( )11 + (-9)
2S = , 10 + 6
2
S = (1 , 8)
mPS =y2 – y1
x2 – x1
= 12-3
=8 – (-4)1 – 4
= -4
y – b = m(x – a)
y – (-4) = -4(x – 4)
y + 4 = -4x + 16
y + 4x = 12
MEDIAN PS
P
R
Q
T
mPQ =y2 – y1
x2 – x1
= 147
=10 – (-4)11 – 4
= 2
ALTITUDE RT
Since perpendicular m1 x m2 = -1,
so mRT = -½
y – b = m(x – a)
y – 6 = -½(x – (-9))
2y - 12 = -x - 9
2y + x = 3
COORDINATES of K
y + 4x = 12 2y + x = 3
1
2
Sub x = 3 into y = -4x + 12 y = -4x3 + 12
1 x 2 2y + 8x = 24 3
3 2- 7x = 21
y = 0 y = -12 + 12
x = 3
P
R
Q
T
S
K
Therefore K is (3 , 0)
A straight line which BISECTS (cuts in half) another straight line & which is PERPENDICULAR to the line.
Perpendicular Bisectors
EXAMPLE 12 Find the equation of the perpendicular bisector of the line PQ, where P(7 , -12) & Q(-3 , 4).
( )x1 + x2
2
y1 + y2
2M = ,
( )7 + (-3)
2M = ,
-12 + 4
2
M = (2 , -4)
y – b = m(x – a)
y – (-4) = ⅝(x – 2)
8y + 32 = 5(x - 2)
8y + 32 = 5x – 10
8y - 5x = -42
Equation ofPerpendicular
Bisector:
mPQ =y2 – y1
x2 – x1
= 16-10
=4 – (-12)-3 – 7
= -85
Since perpendicularm1 x m2 = -1,
so mBIS = 5/8
Centroid
Useful Definitions
Orthocenter
Circumcenter
Point where the 3 Medians intersect!
Point where the 3 Altitudes intersect!
Point where the 3 Perpendicular Bisectors intersect! Centre of a circle joining all three vertex of the triangle!