Answer:
y+3=-4(x-5)
Explanation:
Part A
Given the line:
[tex]y-1=\frac{1}{4}(x-2)[/tex]
We want to find the equation of a perpendicular line that passes through the point (5,-3),
First, determine the slope of the perpendicular line.
Comparing the given line with the slope-point form:
[tex]\begin{gathered} y-y_1=m(x-x_1) \\ \implies\text{Slope},m=\frac{1}{4} \end{gathered}[/tex]
By definition, two lines are perpendicular if the product of their slopes is -1.
Let the slope of the perpendicular line = n
[tex]\begin{gathered} \implies\frac{1}{4}n=-1 \\ n=-4 \end{gathered}[/tex]
Thus, using a slope of -4 and a point (5,-3), we find the equation of the line.
[tex]\begin{gathered} y-y_1=m(x-x_1) \\ y-(-3)=-4(x-5) \\ y+3=-4(x-5) \end{gathered}[/tex]
The equation of the perpendicular line in the slope-point form is:
[tex]y+3=-4(x-5)[/tex]
Part B
In order to graph the line, first, find two points on the line.
When x=0
[tex]\begin{gathered} y+3=-4(0-5) \\ y+3=20 \\ y=20-3=17 \\ \implies(0,17) \end{gathered}[/tex]
When y=1
[tex]\begin{gathered} 1+3=-4(x-5) \\ 4=-4(x-5) \\ \frac{4}{-4}=x-5 \\ -1=x-5 \\ x=5-1=4 \\ \implies(4,1) \end{gathered}[/tex]
Join the points (0,17) and (4,1) as shown in the graph below:
