Systems of linear equations: An equation of a line
Composing the equation of a line
Graphical
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# |
Step 1 | The slope is given, and equal to #5#. |
Step 2 | We enter the slope in the equation of a line #y=a \cdot x+b#. Hence, the equation is of the form : \[y=5\cdot x+b\] in which #b# is a number. |
Step 3 | We enter the point #\rv{-2,1}# in the equation from step 2. This gives us: \[ 1=5\cdot (-2)+b\] |
Step 4 | We solve the equation from step 3 by means of reduction. \[\begin{array}{rcl} 1&=&5\cdot (-2)+b \\ &&\phantom{xxx}\blue{\text{the equation to be solved}}\\ 1&=&-10+b \\ &&\phantom{xxx}\blue{\text{simplified}}\\ 11&=&b \\ &&\phantom{xxx}\blue{\text{both sides minus }-10}\\ \end{array}\] Hence, we find #b=11#. |
Step 5 | We now enter #b=11# in the equation from step 2. We find that the line is given by the equation: \[y= 5\cdot x +11\] |
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