Activity 4.1.1.
Work in pairs
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Randomly select points (e.g., (2,3), (-4,1), (0,-2)) and plot them on a Cartesian plane.
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Exchange your plotted points with another group and discuss.
Section 4.1 Coordinates and Graphs
Coordinates refers to the position of a point on a Cartesian plane. A point is represented as \((x, y),\) where "x" tells you how far to move left or right, and "y" tells you how far to move up or down.
Subsection 4.1.1 Plotting Out Points on a Cartesian Plane
The Cartesian plane consists of two perpendicular axes: the horizontal x-axis and the vertical y-axis, which intersect at the origin \((0, 0).\)
When plotting points on a Cartesian plane:
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If your point is \((3, 4),\) move \(3\) units right (positive x direction) and \(4\) units up (positive y direction).
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If the point is \((-2, -5),\) move \(2\) units left (negative x direction) and \(5\) units down (negative y direction).
\(\textbf{ Key Takeaway }\)
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The axes divide the Cartesian plane into four regions known as quadrants each defined by the x-axis and y-axis and named in an anticlockwise direction:
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First Quadrant (Top Right) – Both x and y coordinates are positive (e.g., (3,2)).
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Second Quadrant (Top Left) – x is negative, y is positive (e.g., (-4,5)).
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Third Quadrant (Bottom Left) – Both x and y coordinates are negative (e.g., (-2,-3)).
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Fourth Quadrant (Bottom Right) – x is positive, y is negative (e.g., (6,-1)).
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Each quadrant helps categorize points based on their signs, making it useful for graphing equations and solving coordinate geometry problems.
The location of points using the perpendicular axes, that is, x and y axis that meet at right angle, is known as cartesian-coordinate system. The system helps in locating places on the earth surface.
Example 4.1.1.
Plot the points \(A(3,2),\, B(-2,4),\, C(0,-1)\) on a Cartesian plane.
Example 4.1.2.
Identify and write down the coordinates of the points on the Cartesian plane below.
Subsection 4.1.2 Drawing a Straight Line Graph Given an Equation
Activity 4.1.2.
Work in groups
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Consider the following equation of a line: \(y = 2x + 1.\)
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Generate a table of values by substituting different x-values: \(x = {-2, -1, 0, 1, 2 }\) in the equation.
| \(x\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(2\) |
| \(y\) | \(\) | \(\) | \(\) | \(\) | \(\) |
\(\textbf{ Key Takeaway }\)
Straight-line graph represents: a constant rate of change between two variables.
The structure of the equation of a straight line is: \(y=mx+c\) where;
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\(m\) is the slope (rate of change)
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\(c\) is the y-intercept (starting value)
Example 4.1.3.
Subsection 4.1.3 Drawing Parallel Lines on a Cartesian-plane
Activity 4.1.3.
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Generate tables plot the points, and draw the lines for the equations
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Compare the lines and discuss why they are parallel.
\(\textbf{ Key Takeaway }\)
Parallel lines are lines that never meet and are always the same distance apart.
Parallel lines have the same direction or slope.
Example 4.1.4.
Solution.
First, generate the table of values for both equations.
| \(y = 2x + 3\) |
| \(y = 2x - 4\) |
| \(x\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(2\) |
| \(y\) | \(-1\) | \(1\) | \(3\) | \(5\) | \(7\) |
| \(x\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(2\) |
| \(y\) | \(-8\) | \(-6\) | \(-4\) | \(-2\) | \(0\) |
Both lines are parallel, clearly shown by their identical slope.
Subsection 4.1.4 Relating the Gradients of Parallel Lines
Activity 4.1.4.
Work in pairs
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Consider the following equations: \(y = x + 5 \text{ and } y = x - 3\)
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Plot the graph of the equations on a cartesian plane.
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Calculate the gradients of the lines and compare them to check whether the lines are parallel.
\(\textbf{ Key Takeaway }\)
We can establish whether two or more lines are parallel by calculating and comparing their gradients.
To calculate the gradient (or slope) of a straight line, we use this formula:
\begin{equation*}
\text{Gradient (}m\text{)} = \frac{\text{change in } y}{\text{change in } x} = \frac{y_2 - y_1}{x_2 - x_1}
\end{equation*}
where \((x_1, y_1)\) and \((x_2, y_2)\) are two distinct points on the line.
Parallel lines have equal gradients.
Example 4.1.5.
Find the gradient of the lines. Are the lines parallel?
Solution.
Calculate the gradient \(m_1\) for the line \(y = 4x +5\) by taking two points in the line: \((-1, 1) \text{ and } (0,5)\)
\begin{equation*}
\text{Gradient (}m_1\text{)} = \frac{5 - 1}{0 - -1}
\end{equation*}
\begin{equation*}
m_1 = \frac{4}{1} = 4.
\end{equation*}
Calculate the gradient \(m_2\) for the second line \(y = 4x - 3\) by taking the points \((0, -3) \text{ and } (1,1)\)
\begin{equation*}
\text{Gradient (}m_2\text{)} = \frac{1 - -3}{1 - 0}
\end{equation*}
\begin{equation*}
m_2 = \frac{4}{1} = 4.
\end{equation*}
Notice that the gradients \(m_1 = m_2 = 4.\) Therefore, parallel lines have the same slope or gradient. You also see that the gradient for the equations is the co-efficient of \(x\) for the equations.
Subsection 4.1.5 Drawing Perpendicular Lines on a Cartesian Plane
Activity 4.1.5.
Work in pairs
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Consider the following equations: \(y = 2x + 1 \text{ and } y = -\frac{1}{2}x + 4.\)
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Generate tables, plot the points, and draw the lines on a Cartesian plane.
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Do the lines intersect? What is the angle of intersection?
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Discuss your findings with other learners.
\(\textbf{ Key Takeaway }\)
Perpendicular lines are lines that intersect at a right angle (90°) for example the x- and y-axes on a Cartesian plane, road intersections.
Example 4.1.6.
Solution.
Subsection 4.1.6 Relating the Gradients of Perpendicular Lines
Activity 4.1.6.
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Consider the following equations: \(y = 5x + 4 \text{ and } y = -\frac{1}{5}x + 1.\)
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Calculate the gradients of the lines and compare them to check whether the lines are perpendicular.
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Multiply the gradients and observe the result.
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Plot the graph of the equations on a cartesian plane.
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Discuss you results with other learners.
\(\textbf{ Key Takeaway }\)
Perpendicular lines have gradients whose product is \(-1.\)
Example 4.1.7.
Find the gradients of \(y = 2x + 4 \text{ and } y = -\frac{1}{2}x + 1\)
Solution.
Generate the table of values for the equations and draw the graph.
| \(y = 2x + 4\) |
| \(y = -\frac{1}{2}x + 1\) |
| \(x\) | \(-2\) | \(0\) | \(1\) |
| \(y\) | \(0\) | \(4\) | \(6\) |
| \(x\) | \(-2\) | \(0\) | \(2\) |
| \(y\) | \(2\) | \(1\) | \(0\) |
Calculate the gradient \(m_1,\)for the line \(y = 2x + 4\) by taking the points \((-2,0), \, (0,4)\)
\begin{equation*}
\text{Gradient (}m_1\text{)} = \frac{4 - 0}{0 - -2}
\end{equation*}
\begin{equation*}
m_1 = \frac{4}{2} = 2.
\end{equation*}
Calculate the gradient for the other line \(y = -\frac{1}{2}x + 1\) by taking the points \((-2,2), \, (0,1)\)
\begin{equation*}
\text{Gradient (}m_2\text{)} = \frac{1 - 2}{0 - -2}
\end{equation*}
\begin{equation*}
m_2 = \frac{-1}{2} = -\frac{1}{2}.
\end{equation*}
Multiply the gradients:
\begin{equation*}
m_1 \times m_2 = 2 \times -\frac{1}{2} = -1.
\end{equation*}
The two line are perpendicular since the product of their gradients is -1.
Subsection 4.1.7 Applying Graphs of Straight Lines in Real Life Situation
Graphs of straight lines help us visualize and solve problems involving constant change. Whether you’re tracking expenses, designing a structure, or analyzing data, linear graphs are powerful tools for making sense of the world.
Example 4.1.8.
A printing shop charges a fixed setup fee of KSh 5 and an additional KSh 2 for each page printed. The total cost y (in KSh) for printing x pages is given by the equation: \(y=2x+5.\) What is the cost of printing:
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\(0\) pages
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\(2\) pages
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If a customer has KSh 65, how many pages can they print? Show your working.
Solution.
First, draw the graph of \(y=2x+5.\)
Since the cost (y) is given as KSh. 65, substitute for y then find x, the number of pages.
\begin{equation*}
65 = 2x + 5
\end{equation*}
\begin{equation*}
65 - 5 = 2x
\end{equation*}
\begin{equation*}
60 = 2x
\end{equation*}
\begin{equation*}
\frac{60}{2} = \frac{2x}{2}
\end{equation*}
\begin{equation*}
x = 30 \text{ pages }
\end{equation*}
