Question

Find the equation for the line that passes through the point (5,-3), and that is perpendicular to the line with the equation y-1=1/4(x-2)

Answer 1

y+3=-4(x-5)

Step-by-step explanation:

Part A

Given the line:

`y-1=(1)/(4)(x-2)`

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:

`\begin{gathered} y-y_1=m(x-x_1) \\ \implies\text{Slope},m=(1)/(4) \end{gathered}`

By definition, two lines are perpendicular if the product of their slopes is -1.

Let the slope of the perpendicular line = n

`\begin{gathered} \implies(1)/(4)n=-1 \\ n=-4 \end{gathered}`

Thus, using a slope of -4 and a point (5,-3), we find the equation of the line.

`\begin{gathered} y-y_1=m(x-x_1) \\ y-(-3)=-4(x-5) \\ y+3=-4(x-5) \end{gathered}`

The equation of the perpendicular line in the slope-point form is:

`y+3=-4(x-5)`

Part B

In order to graph the line, first, find two points on the line.

When x=0

`\begin{gathered} y+3=-4(0-5) \\ y+3=20 \\ y=20-3=17 \\ \implies(0,17) \end{gathered}`

When y=1

`\begin{gathered} 1+3=-4(x-5) \\ 4=-4(x-5) \\ (4)/(-4)=x-5 \\ -1=x-5 \\ x=5-1=4 \\ \implies(4,1) \end{gathered}`

Join the points (0,17) and (4,1) as shown in the graph below: