## Equations of straight lines

BACKGROUND KNOWLEDGE

Co-ordinates and Graphs

To understand co-ordinates and graphs, one must have the knowledge of the Cartesian plane. A Cartesian plane is a plane showing the grid on which points can be located by finding the distances from two lines called the axis. The vertical line labeled y is the y-axis and the horizontal line labeled x is the x-axis as shown below.In the plane, A is the the point(2,3).The first number 2 is referred to as the x-co-ordinate and the second number 3 is referred to as the y-co-ordinate.NB. The x co-ordinate is given first followed by the y co-ordinate.

BACKGROUND KNOWLEDGE

Co-ordinates and Graphs

To understand co-ordinates and graphs, one must have the knowledge of the Cartesian plane. A Cartesian plane is a plane showing the grid on which points can be located by finding the distances from two lines called the axis. The vertical line labeled y is the y-axis and the horizontal line labeled x is the x-axis as shown below.

**Lesson objectives**

#### By the end of the lesson, you should be able to:

Express the equation of a straight line in the form of y=mx+c

State the relationship of gradients of parallel lines.

State the relationship of gradients of perpendicular lines

**Graphical representation of linear
inequalities with one unknown.**

An inequality represents a region with one or more
boundaries.

Consider the inequality y < 4.

To get the boundary line, the inequality symbol (<) is replaced with the
equals (=) symbol. Therefore the boundary line for y<4 is y=4. A
boundary line can be drawn as a broken or a continuous line.

For an equality with the symbol > or <, the boundary line us drawn using
a broken line. This means that points on the line are not included in
the required region.

**Example 1**

Represent y < 4 graphically

*Solution*

Draw the boundary line y = 4 (use a broken line)

Diagram 4

To represent the inequality, choose two points (one above and one below the boundary line) and substitute each point in the inequality. The point that satisfies the inequality is in the wanted region and the point that doesn't satisfy the inequality is in the unwanted region. The unwanted region is always shaded.

Consider points (1,1) and (2,5). Point (1,1) is below the boundary line y=4 and the y-value (1) is less than (4). Therefore point (1,1( is in the wanted region since it satisfies the inequality y<4.

Similarly for point (2,5) the y value is 5 which is greater than 4. This shows that point (2,5) is in the unwanted region since it does not satisfy the inequality.

This means the region above the boundary line is the unwanted region is shaded as shown below.

** **

** **

** **

** **

** **

** **

** **

**Diagram 5**

**Example 2**

Represent

**Solution**

The boundary line is y=2.

For an inequality with the symbols
,
the boundary lines ate drawn using a continuous line. This means that
points on the boundary line are included in the wanted region. Draw the
boundary line y=2 (use a continuous line)

Note;

- The shaded region is the unwanted region
- The points on the boundary satisfy the inequality .
- The points on the line satisfy the inequality .

**Example 3**

Represent
graphically.

*Solution*

The boundary line is x=2 and the wanted region if the one where an x
value is equal to ot greater than 2.

**Example 4**

Represent x<-3 graphically.

*Solution*

The boundary line is x= -3 and the wanted region is the one where an x
value is less than 3.

**Graphical representation of Inequalities
involving two unknowns.**

** Example**

Draw a graph of the inequality 3x + 4y,12.

*Solution*

The boundary line is 3x + 4y = 12. Make a table of values of x and y for
3x + 4y = 12 on the Cartesian plane. (Draw the boundary line using a
broken line)

Plot this on the Cartesian plane.

X 0 4

Y 3 0

Use bright colour to shade and draw line

Substituting (1,1) in the equation 3(1) + 4(1) < 12 we get 7 which is
less than 12 which is below the boundary line. Point (1,1) which is
below the boundary line wanted region. Therefore the region above the
line is shaded.

**Solving Simultaneous Linear Inequalities
Graphically**

To solve simultaneous inequalities draw the inequalities in one graph. The solution are given by points in the wanted region.

*Example*

Draw graph of:

yx
+3, x + y <4 and y0
and show the region that satisfy the inequalities.

**Solution**

The boundary lines are y = x + 3, x + y = 4 and y = 0.

The tables of values are as shown below.

(i) y = x + 3

__x____ -3 0__
__x____ 0 4__

**y** 0 3 **y**
4 0

The following is the required graph.

(Use bright colours to draw graph)

R is out required region

Diagram 10 (Note that the wanted region indicated by letter R.)

**INEQUALITIES FROM INEQUALITIES GRAPHS**

In this lesson, we shall define a region which is
already graphed by determining the inequalities that satisfies the
region.

**Example**

State the inequalities that represent the unshaded region R in each of
the following areas.

Case 1 (+)

The gradient of line 1 is 0. The line cuts the y-axis at 2. Therefore using y = mx + c, the equation of the line of all the points in the line y = 2 is always 2.

Similarly, line 2 cuts x-axis at 3 and the
x-coordinates of all the points on the line is always 3. Therefore the
required line is x = 3.

Considering line 1 (y=2) choose a point (0,0) in the wanted region R.
The value of y is 0. Therefore (0,0) is in the wanted region which has y
values that are less than 2.

Note that, the boundary line is broken. Therefore, the required inequalities is y<2. Considering line 2 (x=3) point x value (1) is less than 3.

Therefore point (1,2) is in the region with x values that are less than 3. Note that the line is a continuous line. Therefore, the inequality is x3

**Step 1**

To find the equations for the boundary lines A,B and C.

Line A

Identify two points on the line

(3,0) and (0,2)

Find the gradient (M) of the line

Let P(x,y) be a general point on the line A. Using the
point (3,0)

Alternatively

Use the equation of strait line y = m x + C

Where the line A passes through (0,2) on the y-axis.

The y-intercept (c) is 2 and the gradient (M) is , using the point P(x,y) and substituting in y = mx + c,

The equation becomes

For line B

The line passes through points (0,2) and (1,3), Point (0,2) is on the y
'axis.

Using y = m x + C (equation of a general straight line) C is 2.

The equation is

= y =
+
2

__Line C__

The line passes through points e.g.

(-1,0), (-1, 1), (-1, 2) etc

The x- value in all the points on the line is -1 so the equation is X = -1

**Step 2**

To get the inequalities

Line A Using (0, 0) which is in the wanted region R. Substitute it in the equation using point (0,0) and after substituting

Therefore 0 < 2,

The inequality is

Line B, y = x + 2(-0.5, 2)

Using the point which is in the required region R

Y = 2, and when substituted in the equation, y = -0.5 + 2 = 1.5

and 2 > 1.5

The inequality is

Y > x + 2

For line C(x = -1) using points (-0.5, 2) and (0,2), the x-value is 2 which is greater than 1. Therefore the inequality is x>-1.

In summary region R is defined by the inequalities

i) y = -?x + 2

ii) y>x + 2

iii) x > -1

Equations of straight lines

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Equations of straight lines

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