What is the equation of the line, in slope-intercept form, that is perpendicular to the line with the equation y=6x−1 and passes through the point (−6,4)?

Respuesta :

Greetings!

Answer:

y = [tex]\frac{-x}{6} + \frac{28}{6}[/tex]

Step-by-step explanation:

First, we must find the slope of the current equation.

This is the number in front of the x.

Seeing as this is 6x, the slope of this line is 6

When finding the slope of a line perpendicular, you need to find the [tex]\frac{-1}{slope}[/tex]

So, in this case it is:

[tex]\frac{-1}{6}[/tex]

The negatives cancel out which leave [tex]\frac{1}{2}[/tex]

So the gradient is [tex]\frac{1}{2}[/tex]

Now, to find the equation of a line, you need to use:

y - y1 = m(x - x1)

Where y1 and x1 are the values in the coordinates (-6 , 4)

So y1  = 4, x1 = -6, and m is [tex]\frac{-1}{6}[/tex]. Plug these values in:

y - 4 = [tex]\frac{-1}{6}(x - -6)[/tex]

We need to get rid of the fraction so we multiply the whole equation by 6 (6 * 1/6 = 1):

6y - 24 = - (x - 2)

The minus outside the bracket means that every value inside the bracket is oposite:

6y - 24 = -x + 2

And now simply move the -24 over to the other side, making it a positive:

6y = -x +2 + 26

Simplify:

6y = -x + 28

Divide all values by 6 to get y:

y = [tex]\frac{-x}{6} + \frac{28}{6}[/tex]

So the equation of the line is y = [tex]\frac{-x}{6} + \frac{28}{6}[/tex]


Hope this helps!