Question

Passing through (7.-2) and perpendicular to the line whose equation is y=1/5x+2.Write an equation for the line in point slope form and in slope intercept form

Answer 1

Point-slope form: y + 2 = -5(x - 7)

Slope-intercept form: y = -5x + 33

Step-by-step explanation:

Two lines are perpendicular if their slope multiplies to -1. So, we can find the slope of our equation as:

`\begin{gathered} m*(1)/(5)=-1 \\ m*(1)/(5)*5=-1*5 \\ m=-5 \end{gathered}`

Because the slope of the line y = 1/5x + 2 is 1/5, the number beside the x.

Now, we write the equation of a line in a point-slope form as:

`y-y_1=m(x-x_1)`

Where m is the slope and (x1, y1) are the coordinates of a point. So, replacing m = -5 and (x1, y1) by (7, -2), we get:

`\begin{gathered} y-(-2)=-5(x-7) \\ y+2=-5(x-7) \end{gathered}`

Now, to write the equation in slope-intercept form, we need to solve for y, so:

`\begin{gathered} y+2=-5(x)-5(-7) \\ y+2=-5x+35 \\ y+2-2=-5x+35-2 \\ y=-5x+33 \end{gathered}`

Therefore, the answers are:

Point-slope form: y + 2 = -5(x - 7)

Slope-intercept form: y = -5x + 33