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}-\frac{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.

Stap 1 The slope #a# is calculated with the formula: \[a=\frac{4-0}{4--2}=-\frac{4}{6}=\frac{2}{3}\]

Stap 2 Hence, the equation is of the form #y=\tfrac{2}{3} \cdot x+b#.

Stap 3 We enter the point #A=\rv{-2,0}# in the equation from step 2. This gives us: \[0=\tfrac{2}{3} \cdot -2 +b\]

Stap 4 We solve the equation from step 3 and find: #b=\tfrac{4}{3}#.

Stap 5 Hence, the equation is #y=\tfrac{2}{3} \cdot x +\tfrac{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{-5,2}# 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 +12 #

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{-5,2}# in the equation from step 2. This gives us: \[ 2=2\cdot (-5)+b\]
Step 4	We solve the equation from step 3 by means of reduction. \[\begin{array}{rcl} 2&=&2\cdot (-5)+b \\ &&\phantom{xxx}\blue{\text{the equation to be solved}}\\ 2&=&-10+b \\ &&\phantom{xxx}\blue{\text{simplified}}\\ 12&=&b \\ &&\phantom{xxx}\blue{\text{both sides minus }-10}\\ \end{array}\] Hence, we find #b=12#.
Step 5	We now enter #b=12# in the equation from step 2. We find that the line is given by the equation: \[y= 2\cdot x +12\]

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