Coordinate Geometry
In this article, we will learn about two topics of coordinate geometry in the Mathematics subject; the Equation of a Straight Line & Geometrical Problems Involving Coordinates.
This article is meant for Secondary 3 students.
Topic Overview:
- Interpret and find the equation of a straight line graph in the form \(y = mx + c\)
- Combined geometric problems involving the use of coordinates
The Equation Of A Straight Line
The equation of the straight line in the graph below is \(y = 4.\)
Evaluation Of A Vertical Line
Here simply, we could say that the equation of the vertical line is \(x = 5.\)
General Equation Of A Straight Line
We know that a non-horizontal and a non-vertical line straight line could be written as
\(y = mx + c\) , where the constant \(m\) is the gradient of the line, and the constant \(c\) is the \(y-intercept\) of the line.
Example:
What is the equation of the following line?
Let us consider two points on the line with the coordinates \((\,2, \,3\,) \)and \((\,6, \,5\,) \)
Gradient of the line:
\(\begin{align*} m &= \frac {(5 −3)}{(6 −2)} \\ \\ &= \frac {2}{4} \\ \\ &= \frac {1}{2} \end{align*}\)
y-intercept: \(c = 2\)
\(\begin{align*} y &= mx + c \\ \\ y &= \frac {1}{2}x +2 \end{align*}\)
Geometrical Problems Involving Coordinates
Example 1:
A straight line passes through the point \((\,1, \,5\,) \) and has a gradient of \(2\). Find the value of the y-intercept.
Solution:
Given: \(m = 2\), find \(c.\)
Given: \(x = 1, \;y = 5\)
\(\begin{align*} y &= mx + c \\ \\ y &= 2x +2 \end{align*}\)
Putting values of \(x = 1, \;y = 5\)
\(\begin{align*} 5 &= 2(1) + c \\ \\ c &= 5-2\\ \\ &= 3 \end{align*}\)
Example 2:
Find the equation of the line passing through each of the following pairs of points.
A) \(A(4, 8)\) and \(B (2, 4)\)
B) \(C (1, 9)\) and \(D (5, -3)\)
Answer:
We will be using the equation \(\begin{align*} m = \frac {y_2 - y_1}{x_2-x_1} \end{align*}\).
For A):
The first step is to find our gradient.
\(\begin{align*} m &= \frac {y_2 - y_1}{x_2-x_1} \\ \\ &= \frac {8 - 4}{4-2} \\ \\ &= \frac {4}{2} \end{align*}\)
Now, we will be using the equation, \(y = mx + c\).
\(y = 2x + c\)
Substituting \(x = 4 \) and \(y = 8\).
\(\begin{align*} 8 &= 2 (4) + c \\ \\ 8 &= 8 + c \\ \\ c &= 0 \end{align*}\)
So, the equation of the line will be \(y = 2x\).
For B):
The first step is to find our gradient.
\(\begin{align*} m &= \frac {y_2 - y_1}{x_2-x_1} \\ \\ &= \frac {9 - (-3)}{1-5} \\ \\ &= \frac {9+3}{-4} \\ \\ &= \frac {12}{-4} \\ \\ &= -3 \end{align*}\)
Now we will use \(y = mx + c\).
\(y = -3x + c\)
Substituting, \(x = 1\) and \(y = 9\)
\(\begin{align*} 9 &= -3 (\,1\,) + c \\ \\ 9 &= -3 + c \\ \\ c &= 9 + 3 \\ \\ &= 12 \end{align*}\)
So, the equation of the line will be \(\begin{align*} y = - 3x + 12 \end{align*}\).
Example 3:
Find the equation of a straight line by making \(y\) the subject of the formula.
Given the line \(4y-2x=7\),
A) Make \(y\) the subject of the formula
\(4y-2x=7\), and
B) Hence, find out the
i) gradient, and
ii) the coordinates of the \(y\)-intercept of the line
Answer:
A)
\(4y-2x=7\)
We will keep only the y term on the left-hand side, and the rest will go on the right-hand side.
\(\begin{align*} 4y &=2x+ 7 \\ \\ y &=\frac {2x+ 7}{4} \\ \\ y &= \frac{7}{4} +\frac{2x}{4} \\ \\ y &= 1\frac{3}{4} +\frac{1}{2}x \\ \\ y &=\frac{1}{2}x + 1\frac{3}{4} \end{align*}\)
B)
Hence, here indicates that we need to use the answer from part A. Here, we have to find out the y-intercept using the answer from part A.
\(\begin{align*} y &=\frac{1}{2}x + 1\frac{3}{4} \\ \\ y &= mx +c \end{align*}\)
So, by comparing the above two equations, we can say that \(m = \frac {1}{2}\) and the \(y-intercept = 1\frac{3}{4}\).
Thus, the coordinate of the \(y-intercept\) will be \((0, 1\frac{3}{4})\).
i) Gradient \(= \frac {1}{2}\)
ii) Coordinate of the y-intercept will be \((0, 1\frac{3}{4})\) or \((0, 1.75)\).
Example 4:
A line passes through the points \(A (-2, -4)\) and \(B (4, 11)\). In another line, \(P\) has an equation \(2y+6= 5x\).
Show how you can tell that line \(P\) does not intersect line \(AB\).
Answer:
We have to show that line \(P\) is parallel to line \(AB\), and how can we show that?
We need to show that they have the same gradient.
We need to find the gradient of both the lines.
For line \(AB\):
\(\begin{align*} m &= \frac {y_2 - y_1}{x_2-x_1} \\ \\ &= \frac {-4 - 11}{-2-4} \\ \\ &= \frac {-15}{-6} \\ \\ &= 2.5 \end{align*}\)
By manipulating the line equation for \(P\),
\(\begin{align*} 2y+6 &= 5x \\ \\ \implies\qquad 2y &= 5x-6 \\ \\ \implies\qquad\;\; y &=\frac {5x-6}{2} \\ \\ \implies\qquad\;\; y &=\frac {5x}{2}- \frac{6}{2} \\ \\ \implies\qquad\;\; y &=2.5x-3 \end{align*}\)
So, the gradient of line \(P\) is equal to \(2.5\).
Since, the gradient of line \(AB=\) gradient of line \(P=2.5\),
Line \(AB\) is parallel to line \(P.\)
Therefore, they do not intersect.
Conclusion
In this article, we learned how to interpret and find the equation of a straight line graph in the form \(y = mx + c\). Furthermore, we also solved combined geometric problems involving the use of coordinates and learned how to tackle different pattern questions of such type.
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Continue Learning | |
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Further Trigonometry | Quadratic Equations And Functions |
Linear Inequalities | Laws of Indices |
Coordinate Geometry | Graphs Of Functions And Graphical Solution |
Applications Of Trigonometry |