Composing the equation of a line

Systems of linear equations: An equation of a line

Composing the equation of a line

Graphical

geogebra plaatje

Composing a linear equation through two points

	Procedure	Example
	Compose an equation of the line through two points $A=\rv{\blue{x_A}, \blue{y_A}}$ and $B=\rv{\green{x_B}, \green{y_B}}$ .	Assume that $A=\rv{\blue2, \blue5}$ and $B=\rv{\green4,\green2}$ .
Step 1	Determine the slope $a$ with help of the formula: $a=\frac{\green{y_B}-\blue{y_A}}{\green{x_B}-\blue{x_A}}$	$\begin{array}{rcl}a&=&\dfrac{\green{y_B}-\blue{y_A}}{\green{x_B}-\blue{x_A}}\\ &=&\dfrac{\green2-\blue5}{\green4-\blue2}\\ &=&-\dfrac{3}{2}\end{array}$
Step 2	The equation is of the form $y=a \cdot x+b$ with $a$ from step 1 and $b$ as an undefined number.	$y=-\dfrac{3}{2} \cdot x+b$
Step 3	Enter the point $A=\rv{\blue{x_A},\blue{y_A}}$ in the equation from step 2: $\blue{y_A}=a \cdot \blue{x_A}+b$	$\blue5=-\dfrac{3}{2} \cdot \blue2 +b$
Step 4	Solve the equation from step 3 for unknown $b$ .	$\begin{array}{rcl}-\dfrac{3}{2} \cdot \blue{2} +b&=&\blue{5} \\ -3+b&=&5 \\ b &=&8 \end{array}$
Step 5	Use the value of $b$ : $y=a \cdot x+b$ found in step 4.	$y=-\dfrac{3}{2} \cdot x +8$

Horizontal line

An exception to this procedure is the horizontal line. This line can be recognized because the $y$ -coordinates of the two points are equal. Hence, this is a line through the points $A=\rv{x_A,\orange c}$ and $B=\rv{x_B,\orange c}$ . This line is equal to $y=\orange c$ .

Example

The line through the points $A=\rv{2,\orange8}$ and $B=\rv{6, \orange8}$ satisfies the equation $y=\orange 8$

Vertical line

An exception to this procedure is the vertical line. This line can be recognized because the $x$ -coordinates of the two points are equal. Hence, this is a line through the points $A=\rv{\orange c, y_A}$ and $B=\rv{\orange c, y_B}$ . This line is equal to $x=\orange c$ .

Example

The line through the points $A=\rv{\orange8, 2}$ and $B=\rv{\orange8,6}$ satisfies the equation $x=\orange 8$

Composing an equation with a graphWith the help of this procedure, we can also compose an equation with the graph of a line. We can do this by reading two convenient points from the graph and following the procedure with these points.

Example

We will compose an equation of the form $y=a \cdot x +b$ for the line below.

We will first pick two convenient points, for example, $A=\rv{-2,0}$ and $B=\rv{4,4}$ . Next, we follow the procedure.

Step 1 The slope $a$ is calculated with the formula: $a=\dfrac{4-0}{4--2}=-\dfrac{4}{6}=\dfrac{2}{3}$

Step 2 Hence, the equation is of the form $y=\dfrac{2}{3} \cdot x+b$ .

Step 3 We enter the point $A=\rv{-2,0}$ in the equation from step 2. This gives us: $0=\dfrac{2}{3} \cdot -2 +b$

Step 4 We solve the equation from step 3 and find: $b=\dfrac{4}{3}$ .

Stap 5 Hence, the equation is $y=\dfrac{2}{3} \cdot x +\dfrac{4}{3}$ .

Given slopeTo compose an equation of a line, it is sufficient to know the slope and one point on the graph. We then actually start with step 2 of the procedure and enter the known slope right away.

Determine an equation for the straight line that passes through $\rv{2,1}$ and has a slope of $2$ .

Write your answer in the form of $y=a\cdot x+b$ with suitable numbers $a$ and $b$ and with unknowns $x$ and $y$ .

$y= 2\cdot x -3$

Step 1	The slope is given, and equal to $2$ .
Step 2	We enter the slope in the equation of a line $y=a \cdot x+b$ . Hence, the equation is of the form: $y=2\cdot x+b$ in which $b$ is a number.
Step 3	We enter the point $\rv{2,1}$ in the equation from step 2. This gives us: $1=2\cdot 2+b$
Step 4	We solve the equation from step 3 by means of reduction. $\begin{array}{rcl} 1&=&2\cdot 2+b \\ &&\phantom{xxx}\blue{\text{the equation to be solved}}\\ 1&=&4+b \\ &&\phantom{xxx}\blue{\text{simplified}}\\ -3&=&b \\ &&\phantom{xxx}\blue{\text{subtracted } 4 \text{ to both sides }}\\ \end{array}$ Hence, we find $b=-3$ .
Step 5	We now enter $b=-3$ in the equation from step 2. We find that the line is given by the equation: $y= 2\cdot x -3$

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