Study S3 Mathematics Coordinate Geometry - Geniebook

# 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.

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 $$(\,6, \,5\,)$$

\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 $$D (5, -3)$$

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

ii) the coordinates of the $$y$$-intercept of the line

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 $$AB$$

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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