Question

What is an equation of the line that passes through the point

(-6, – 7) and is perpendicular to the line 6x + 5y = 30?
HELP!!!

Answer 1

(y+7)=5/6(x+6)

Explanation:

For this problem we need to use Point-Slope Form to write an equation.

We already have the point, (-6,-7), so all we need now is the slope, which we know is perpendicular to the line 6x+5y=30.

First, let's find the slope of 6x+5y=30 by converting it into Slope-Intercept Form, which is y=mx+b. m is the slope.

6x+5y=30

5y=-6x+30

y=-6/5x+6

So the slope of this line is -6/5, but we need the slope of the line is perpendicular to this line.

When a line is perpendicular to another, their slopes are negative reciprocals to one another, or opposite reciprocals. For example, a line that is perpendicular to another line that has a slope of 3/4 will have a slope of -4/3. Flip the numerator and denominator to find the reciprocal and add a negative.

x

So the negative reciprocal of -6/5 is 5/6, so that is our slope.

Now we can assemble our equation, here is the Point-Slope Form:

`(y -y_(1)) =m(x-x_(1))`

again, m is our slope.

`y_(1)` and
`x_(1)` represent the coordinates of the point that this line passes through, (-6, -7).

Let's plug in the y, x, and m values:

(y-(-7))=5/6(x-(-6))

Simplify:

(y+7)=5/6(x+6)

That is our final answer.