Question

What is the equation of the line perpendicular to `y=-15/7x` that passes through `(-1,7)`?

Answer 1

Point-slope form: `y-7=7/15(x+1)`

Slope-intercept form: `y=7/15x+112/15`

Explanation:

The slope of a perpendicular line is the negative reciprocal of the original slope. In this case, the perpendicular slope of `-15/7` is `7/15`. The product of two perpendicular slopes is `-1`.

`-15/7xx7/15=-1`

With the slope and one point, you can write a linear equation in point-slope form:

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

where:

`m` is the slope, and `(x_1,y_1)` is the given point.

Plug in the known values.

`y-7=7/15(x-(-1))`

Simplify.

`y-7=7/15(x+1)`

You can convert the point-slope form to slope-intercept form by solving for `y`. `(y=mx+b)`

`y=7/15x+7/15+7`

Multiply `7` by `15/15` to get an equivalent fraction with the denominator `15`.

`y=7/15x+7/15+7xx15/15`

`y=7/15x+7/15+105/15`

`y=7/15x+112/15` `larr` slope-intercept form

graph{y-7=7/15(x+1) [-10.04, 9.96, 1.44, 11.44]}