Find the line perpendicular to y= 3x-5 that passes through (6,11).

Question

I really don't know how to solve this! Please help!

Answer 1

First, lines that are perpendicular have slopes that are negative reciprocals of one another. So, this means that the new line will have a slope that is the negative reciprocal of 3.

3 is a whole number, but technically it is over 1. So, 3/1 becomes 1/3 if we flip it, then take the negative, this is the new slope. -1/3.

Now, take the new slope and the given point and plug into point-slope form.

y-11 = -1/3(x - 6)

y- 11= -1/3x + 2 Use distributive property and you can convert to slope-intercept form if you choose.

y-11+11 = -1/3x + 2+11

y= -1/3x + 13

Answer 2

Find the line perpendicular to y= 3x-5 that passes through (6,11).

=>

slope of line perpendicular to given line = -1/3

y= mx +b -----------slope intercept form

y=(-1/3)x + b

to find b , substitute (6,11) in equation

11 =(-1/3)6 +b

11 =-2+b

13 = b

y= (-1/3)x + 13