Explanation:
We are required to draw the graph of the line that is parallel to y-3=1/3(x+2) and goes through the point (1, 7).
Given the equation of the line:
[tex]y-3=\frac{1}{3}(x+2)[/tex]
Compare the equation with the slope-point form of a line:
[tex]$y-y_1=m(x-x_1)$[/tex]
• The slope of the line, m=1/3
,
• In addition, the line goes through the point (1,7)
Substitute these values into the point-slope form given above:
[tex]y-7=\frac{1}{3}(x-1)[/tex]
Finally, graph the line by looking for another point in addition to point (1,7):
When x=-2
[tex]\begin{gathered} y-7=\frac{1}{3}(x-1) \\ y-7=\frac{1}{3}(-2-1) \\ y-7=\frac{1}{3}(-3) \\ y-7=-1 \\ y=-1+7 \\ y=6 \\ \implies(-2,6) \end{gathered}[/tex]
Join the points (1, 7) and (-2, 6) to plot the line.
Answer:
The graph showing the two points is attached below:
Note:
For comparison purposes and to show that the two lines are parallel, the other graph is added below:
