Question

What is the equation of the line that passes through the point 2,6 and is perpendicular to the line 3x-y=12

Answer 1

We are given 3x - y = 12

This means y = 3x - 12

The perpendicular line is:

$y - 6 = - \frac{1}{3}(x - 2) \\ \\ this \: means \: y - 6 = - \frac{1}{3}x + \frac{2}{3} \: and \\ \\ y = - \frac{1}{3}x + \frac{20}{3} is \: the \: perpendcular \: line$

Answer 2

Answer:

y-6=-1/3(x-2)

Step-by-step explanation:

We first need to find the slope of the line 3x-y=12

Solve for y by subtracting 3x from each side

-3x-3x-y = -3x+12

-y = -3x+12

Divide by -1

-y/-1 = -3x/-1 +12/-1

y = 3x-12

The slope is 3

For parallel lines the slopes are negative reciprocals

Flip the slope and negate it

3 becomes 1/3 and then -1/3

So the slope of the new line is -1/3

We have the slope -1/3 and a point (2,6)  we can use point slope form of a line to determine the equation

y-y1 = m(x-x1)

y-6=-1/3(x-2)

This is in point slope form