Write an equation of a line passing through (3,1) that is perpendicular to y=1/3x-5

Question

I need help 

Answer 1

If you have the slope of a line (m), and want to get the slope of a line perpendicular to it, you need to take the negative reciprocal of m [-1/m].

 

Your line: y = 1/3 x - 5

It is in the form, y = mx + b, where m is its slope. Therefore, its slope is 1/3. The slope of the perpendicular line is -1/[1/3], or -3.

 

So far, your new line's equation is y = -3x + b. Now, you need to find b, the y-intercept. Since you know a point the new line passes through [(3, 1)], you have an x [3] and a y [1]. Substitute in the new line's equation:

 

1 = -3(3) + b, solve for b

1 = -9 + b

10 = b

 

The new line's equation is y = -3x + 10.

Answer 2

When two lines are perpendicular to each other the product of their slope

 

is equal to - 1

 

let the slope of the line be m then

 

m* (1/3) = -1

 

m = -3

 

now all we need to do is write the standard eq for a line in 

 

y = mx + b form

 

(y - 1) = -3 ( x - 3)

 

y = -3x + 9  + 1

 

y = -3x + 10

 

thats the equation of your line