Question

Which equation in point -slope form describes the line that passes through the point (4,-5) and is perpendicular to the line represented by -7x+2y=14.

Answer 1

`y+5=-(2)/(7) (x-4)`

Explanation:

Hi there!

What we need to know:

• Linear equations are typically organized in slope-intercept form:
  `y=mx+b` where m is the slope and b is the y-intercept (the value of y when the line crosses the y-axis)
• Point-slope form:
  `y-y_1=m(x-x_1)` where the given point is
  `(x_1,y_1)` and m is the slope
• Perpendicular lines always have slopes that are negative reciprocals (ex. 2 and -1/2, 3/4 and -4/3)

1) Determine the slope (m)

`-7x+2y=14`

First, rearrange this given equation into slope-intercept form (
`y=mx+b`) so we can easily find the slope.

Add 7x to both sides to isolate 2y

`-7x+2y+7x=7x+14\\2y=7x+14`

Divide both sides by 2 to isolate y

`y=(7)/(2) x+7`

Now, we can identify clearly that the slope of this line is
`(7)/(2)`. Because perpendicular lines have slopes that are negative reciprocals, we know that the slope of the line we're currently solving for will have a slope of
`-(2)/(7)`.

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

2) Plug all necessary values into
`y-y_1=m(x-x_1)`

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

We know that
`m=-(2)/(7)` and that the given point is (4,-5). Plug the slope into the equation

`y-y_1=-(2)/(7) (x-x_1)`

Plug the point into the equation

`y-(-5)=-(2)/(7) (x-4)\\y+5=-(2)/(7) (x-4)`

I hope this helps!