Read the coordinates of a point

Coordinates of a points β are the values of π₯ and π¦ enclosed by the bracket which are used to describe the position of a point in the plane

The plane used is called π₯π¦ β plane and it has two axis; horizontal axis known as π₯ β axis and; vertical axis known as π¦ β axis

A Point Given its Coordinates

Plot a point given its coordinates

The

*x*-coordinate (

*always*comes first. The first number (the first coordinate) is

*always*on the horizontal axis.

A Point on the Coordinates

Locate a point on the coordinates

The location of (2,5) is shown on the coordinate grid below. The

*x*-coordinate is 2. The*y*-coordinate is 5. To locate (2,5), move 2 units to the right on the*x*-axis and 5 units up on the*y*-axis.The order in which you write

*x*β and*y*-coordinates in an ordered pair is very important. Th e*x*-coordinate always comes first, followed by the*y*-coordinate. As you can see in the coordinate grid below, the ordered pairs (3,4) and (4,3) refer to two different points!**Gradient (Slope) of a Line**

The Gradient of a Line Given Two Points

Calculate the gradient of a line given two points

Gradient

or slope of a line β is defined as the measure of steepness of the

line. When using coordinates, gradient is defined as change in π¦ to the

change in π₯.

or slope of a line β is defined as the measure of steepness of the

line. When using coordinates, gradient is defined as change in π¦ to the

change in π₯.

Consider two points π΄ (π₯

_{1}, π¦_{1)}and (π΅ π₯_{2}, π¦_{2)}, the slope between the two points is given by:Example 1

Find the gradient of the lines joining:

- (5, 1) and (2,β2)
- (4,β2) and (β1, 0)
- (β2,β3) and (β4,β7)

**Solution**

Example 2

- The line joining (2,β3) and (π, 5) has gradient β2. Find π
- Find the value of π if the line joining the points (β5,β3) and (6,π) has a slope ofΒ½

**Solution**

**Equation of a Line**

The Equations of a Line Given the Coordinates of Two Points on a Line

Find the equations of a 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 π¦ β intercept (at x = 0) or π₯ β intercept ( at y=0)
- The gradient and a point on the line
- Since only one point is given, then

- Two points on the line

Example 3

Find the equation of the line with the following

- Gradient 2 and π¦ β intercept β4
- Gradient β2β3and passing through the point (2, 4)
- Passing through the points (3, 4) and (4, 5)

**Solution**

Divide by the negative sign, (β), throughout the equation

β΄The equation of the line is 2π₯ + 3π¦ β 16 = 0

The equation of a line can be expressed in two forms

- ππ₯ + ππ¦ + π = 0 and
- π¦ = ππ₯ + π

Consider the equation of the form π¦ = ππ₯ + π

π = Gradient of the line

Example 4

Find the gradient of the following lines

- 2π¦ = 5π₯ + 1
- 2π₯ + 3π¦ = 5
- π₯ + π¦ = 3

**Solution**

Intercepts

The line of the form π¦ = ππ₯ + π, crosses the π¦ β ππ₯ππ when π₯ = 0 and also crosses π₯ β ππ₯ππ when π¦ = 0

See the figure below

Therefore

- to get π₯ β intercept, let π¦ = 0 and
- to get π¦ β intercept, let π₯ = 0

From the line, π¦ = ππ₯ + π

π¦ β intercept, let π₯ = 0

π¦ = π 0 + π = 0 + π = π

π¦ β intercept = c

Therefore, in the equation of the form π¦ = ππ₯ + π, π is the gradient and π is the π¦ β intercept

Example 5

Find the π¦ β intercepts of the following lines

**Solution**

**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:

- By using intercepts
- By using the table of values

Example 6

Sketch the graph of π¦ = 2π₯ β 1

**Solution**

The Graph of a Linear Equation

Draw the graph of a linear equation

By using the table of values

**Simultaneous Equations**

Linear Simultaneous Equations Graphically

Solve linear simultaneous equations graphically

Use

the intercepts to plot the straight lines of the simultaneous

equations. The point where the two lines cross each other is the

solution to the simultaneous equations

the intercepts to plot the straight lines of the simultaneous

equations. The point where the two lines cross each other is the

solution to the simultaneous equations

Example 7

Solve the following simultaneous equations by graphical method

**Solution**

Consider: π₯ + π¦ = 4

If π₯ = 0, 0 + π¦ = 4 π¦ = 4

If π¦ = 0, π₯ + 0 = 4 π₯ = 4

Draw a straight line through the points 0, 4 and 4, 0 on the π₯π¦ β plane

Consider: 2π₯ β π¦ = 2

If π₯ = 0, 0 β π¦ = 2 π¦ = β2

If π¦ = 0, 2π₯ β 0 = 2 π₯ = 1

Draw a straight line through the points (0,β2) and (1, 0) on the π₯π¦ β plane

From the graph above the two lines meet at the point 2, 2 , therefore π₯ = 2 πππ π¦ = 2

Tags: MATHEMATICS

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