Question

What is the equation of the line perpendicular to `y=-2/7x` that passes through `(-2,5)`?

Answer 1

`y-5 = 7/2(x+2)` Equation in point-slope form.
`y=7/2x+12` Equation of the line in slope-intercept form

Explanation:

To find the equation of the line perpendicular to the given line.

Step 1: Find the slope of the given line.

Step 2: Take the slope's negative reciprocal to find the slope of perpendicular.

Step 3: Use the given point and the slope use the Point-Slope form to find the equation of line.

Let us write our given line and go through the steps one by one.

`y=-2/7x`

Step 1: Finding the slope of `y=-2/7x`
This is of the form `y=mx+b` where `m` is the slope.

Slope of the given line is `-2/7`

Step 2: The slope of perpendicular is the negative reciprocal of the given slope.

`m= -1/(-2/7)`
`m=7/2`

Step 3: Use the slope `m=7/2` and the point #(-2,5) to find the equation of the line in the Point-Slope form.

Equation of line in Point-slope form when slope `m` and a point `(x_1,y_1)` is `y-y_1 = m(x-x_1)`

`y-5 = 7/2(x+2)` Solution in point-slope form.

Simplifying we can get
`y-5 = 7/2x+7` using distributive propertly
`y = 7/2x +7+5` adding `5` both sides

`y=7/2x+12` Equation of the line in slope-intercept form