Answer:
[tex]\textsf{a)} \quad y=-4x+19[/tex]
[tex]\textsf{b)} \quad y=\dfrac{1}{4}x+2[/tex]
Step-by-step explanation:
From inspection of the graph, two points on the line are:
Substitute these points in the slope formula to find the slope of the line:
[tex]\implies \textsf{slope}\:(m)=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{2-6}{2-1}=\dfrac{-4}{1}=-4[/tex]
Part (a)
Lines that are parallel have the same slope.
Therefore, a line parallel to the given line has a slope of -4.
To write an equation of the parallel line, substitute the found slope and point (4, 3) into the point-slope formula and rearrange:
[tex]\implies y-y_1=m(x-x_1)[/tex]
[tex]\implies y-3=-4(x-4)[/tex]
[tex]\implies y-3=-4x+16[/tex]
[tex]\implies y=-4x+19[/tex]
Part (b)
If two lines are perpendicular to each other, their slopes are negative reciprocals.
Therefore, a line perpendicular to the given line has a slope of ¹/₄.
To write an equation of the parallel line, substitute the found slope and point (4, 3) into the point-slope formula and rearrange:
[tex]\implies y-y_1=m(x-x_1)[/tex]
[tex]\implies y-3=\dfrac{1}{4}(x-4)[/tex]
[tex]\implies y-3=\dfrac{1}{4}x-1[/tex]
[tex]\implies y=\dfrac{1}{4}x+2[/tex]
