Basic Mathematics Form One Notes – Coordinate Geometry
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Coordinates of a Point
The Coordinates of a Point
Read the coordinates of a point
Coordinates of a points ~ are the values of x and y enclosed by the bracket which are used to
describe the position of a point in the plane
The plane used is called xy ~ plane and it has two axis; horizontal axis known as x ~ axis and;
vertical axis known as y ~ axis,
A Point Given its Coordinates
Plot a point given its coordinates
Suppose you were told to locate (5, 2) on the plane. Where would you look? To understand the
meaning of (5, 2), you have to know the following rule: Thex-coordinate (alwayscomes first. The
first number (the first coordinate) isafwayson the horizontal axis
aa
y
5
4
F)
So, for the point (5, 2), you would stat at
the ‘origin’ the spot where the axes iy
cross: ¥
4T123 456?
1
y
5
4
5)
{then count over to “five” on the x-axis 2
1 x
aT1es 456?
Mt
a
y
5
4
3
then count up to “wo”, moving parallel 2
to the y-axis:
y x
TTies ape?
4
2
y
5
4
3
land then draw in the dot: 2
¥ x
TT1es 456?
oT
2
A Point on the Coordinates
Locate a point on the coordinates
5
The location of (2,5) is shown on the coordinate grid below. Thex-coordinate is 2. They-
coordinate is 5. To locate (2,5), move 2 units to the right on thex-axis and 5 units up on they-axis.
6
5 ae
4
3
2
1
0 x
0123 45 6
The order in which you writex- andy-coordinates in an ordered pair is very important. Th ex-
coordinate always comes first, followed by they-coordinate. As you can see in the coordinate
arid below, the ordered pairs (3,4) and (4,3) refer to two different points!
116
y
6
5
4 ° |
3 *
2
1
0 :
o12z3 456
Gradient (Slope) of a Line
The Gradient of a Line Given Two Points
Calculate the gradient of a line given two points
coordinates, gradient is defined as change in y to the change in x.
change in x
rea
on
X41 — %2|
a7
Example 1
Find the gradient of the lines joining:
a(S, 1) and (2-2)
b. @-2)and (-1, 0)
© (2-3)and-4-7)
Solution
(a) (5,1) and (2,-2)
Ye _-2@=1_=3_
gradient = gag cago
(b) (4,-2) and (—1,0)
_Ya-y_O–2_ 2 1
gradient = ge ae7 3
(c) (-2,-3) and (-4,-7)
_Ya7yy _-7- -3_-74+3_-4_
eradient = = goa aay 7227?
Example 2
a. The line joining (2,~3) and (k, 5) has gradient ~2. Find k
b. Find the value of m if the line joining the points (~5,~3) and (6,m) has a slope of”
Solution
a8
(a) Given (2,-3) and (k,5)
gradient = 22%
2= 5–3
“k=?
-2(k-2)=54+3
-2k+4=8
—2k=8-4
-2k=4
k= a
-s-
+ The value of k is —2
(b) Given (—5,~3) and (6,m)
gradient = 22—
2 –
1_m–3
2° 6–5
1_m+3
2° 645
1_m+3
2° 11
2(m+3)=11
2m+6=11
2m=11-6
2m=5
5
maa
The value of kis =
Equation ofa Line
a9
‘The Equations of a Line Given the Coordinates of Two Points on a Line
Find the equations ofa line given the coordinates of two points on a line
The equation of a straight line can be determined if one of the following is given:~
+The gradient and the y intercept (at x = 0) or x ~ intercept ( at y~0)
+The gradient and a point on the line
+ Since only one point is given, then
Gem oe
jeradient = —*
+ Two points on the line
Example 3
FFind the equation of the line with the following
a. Gradient 2 and y ~ intercept ~4
b. Gradient ~24and passing through the point (2, 4)
©. Passing through the points (3, 4) and (4, 5)
Solution
320
(a) Given m=2 and c=-4
yemxte
ys2x-4
(b) Recall
avery
gradient = 7
2_y-4
3 x-2
-2(x-2)=37-4)
-2x +4 = 3y-12
-2x+4-3y+12=0
-2x -3y+16=0
Divide by the negative sign, (~), throughout the equation
«The equation ofthe line is 2x + 3y- 16 =0
aa
(c) Recall
Ya-Ys
adient = 724
a x2 %
aS4_1_,
m=4-3°17
Then
gradient = 2—TM*
xx
y-4
1-2-4
x3
x-3=y-4
x-3-y+4=0
x-y+1=0
The equation of the line is xy +1=0
The equation ofa line can be expressed in two forms
a ax+by+e=Oand
b-y=mr+e
Consider the equation ofthe form y = mx + €
m= Gradient ofthe line
Example 4
Find the gradient ofthe following lines
a dy=Sx+l
b. dxt3y=5
e xtys3
Solution
Fray
(a) Express in the form of y = mx + ¢
Divide by both sides
Sx+1 5 1
ya QA atta
5 1
yagrts
{b) Express in the form of y = mx +c
Divide by both sides
2x+3y=5
3y =5-2x
3y=-2x+5
—2x+5 2 5
yay att
2 5
yongrte
Gradient = 2
() x+y=3
Express in the form of y = mx +c
y=3-x
y=-x4+3
+ Gradient = —1
Intercepts
when y =0
See the figure below
1
yaxis
of ®y=mxte
x— axis
ato get x~ineroept, let y= O and
b —togety~ intercept, let x= 0
From te line, y= max+¢
‘y~imtercept, let x= 0
y=m0+ce=0+e=¢
y~ intercept =e
Therefore, in the equation of the form y= mx +c, m is the gradient and cis the y~ intercept
Example 5
Find the y ~ intercepts ofthe folowing lines
(a) y= 3×45
() y=-3×42
() 3y=2×41
Solution
na
(a) y= 3×45
Compare with y = mx +c
y— intercept = ¢ = 5
sy — intercept is 5
=-tx4?
(b) y=-3x+?
2
y~ intercept = 5
() 3y=2×41
Express in the form of y = mx +c
Divide by 3 both sides
a+ 211
ae ee
yey 343
wey
ygtty
inte rt 1
~ intercept = =
y Pp 3
Graphs of Linear Equations
The Table of Value
Form the table of value
‘The graph of a straight line can be drawn by using two methods
a. Byusing intercepts
b. _Byusing the table of values
Example 6
Sketch the graph of y =2x~ 1
Solution
25
By using intercepts
y— intercept, let x = 0
y=2(0)-1
y=0-1
ye-1
x— intercept, let y = 0
O=2x-1
2x-1=0
2x=1
“1
at
The coordinates are (,0) and (0,-1)
Then show the straight line through the point (2,0) and (0, -1) on the xy ~plane.
The Graph of a Linear Equation
Draw the graph ofa linear equation
By using the table of values
126
[x]-1T 0 [i[2]3]
=
gs
ey
3
2
4
°
pt eee eee eeeee |
wanis
3
“4
Simultaneous Equations
Linear Simultaneous Equations Graphically
Solve linear simultaneous equations graphically
Use the intercepts to plot the straight lines ofthe simultaneous equations. The point where the
two lines cross each other isthe solution tothe simultaneous equations
Example 7
Solve the following simultaneous equations by graphical method
xty=4
2x-y=2
Solution
Consider: x+y =4
Ifx=0,0+y=4y=4
Ify=0,x+0=4x=4
a7
Draw a straight line through the points 0, 4 and 4, 0 on the xy ~ plane
Consider: 2x-y=2
Ifx=0,0-y=2y=-2
Ify=0,2x-0=2x=1
Draw a straight line through the points (0,~2) and (1, 0) on the xy ~ plane
y
Qx-y=2
x
7654321% 12345 –
4 xty=4
From the graph above the two lines meet atthe point 2, 2, therefore x= 2.and y=2
128