OpenStudy (anonymous):

What is the equation of the line that passes through the point (2, 3) and is perpendicular to the line y = –x – 3?

what is the slope of line y = -x-3 ?

OpenStudy (anonymous):

Would it be 2/3?

OpenStudy (agent47):

OpenStudy (anonymous):

-3?

OpenStudy (agent47):

What is the slope of y=mx+b? Can you google it and tell me what it is?

OpenStudy (anonymous):

Oh the slope is -1

OpenStudy (agent47):

yes. Now if the slope of the original line is M, what is the slope of the line perpendicular to it?

OpenStudy (anonymous):

It would be 1 since its the reciprocal of it right?

OpenStudy (nikato):

To find the slope of a line perpendicular, u find the negative reciprocal of the original slope. So in ur problem, to find the slope of the perpendicular line, u will find the negative reciprocal of -1 which is ....

OpenStudy (anonymous):

1

OpenStudy (agent47):

Correct Algebra, the slope of the perpendicular line is the negative inverse of the original slope.

OpenStudy (agent47):

So you now know that the line passes through (2,3) and has slope of 1.

OpenStudy (the_fizicx99):

It would actually be 1/1 which is equal to 1

OpenStudy (nikato):

Now u can just plug that and the coordinates of the new line which is given in ur problem in the point slope equation

OpenStudy (agent47):

use the following formula: y-y1 = M(x-x1) where (2,3) is (x1, y1).

OpenStudy (the_fizicx99):

Really all you wanted to use from y = -x -3 is the slope, Then you would plug in y-y1=m(x-x1) (2,3) which is: y -2=1(x-3) can you solve now, the first step would be to apply distributive property to the 1(x-3)

OpenStudy (anonymous):

So would it be (x+1)?

OpenStudy (the_fizicx99):

You're making a new equation, and putting it into y = mx + b form

OpenStudy (the_fizicx99):

What is 1 * x and 1 *(-3) @AlgebraOhNo

OpenStudy (anonymous):

x and -3

OpenStudy (nurali):

1st, find the slope, m1, of y = –x – 3 by putting it in slope-intercept form. (That means solve for y) y = –x – 3 m1 = -1 The slope of lines perpendicular is the negative inverse of m1, = 1 Use y = mx + b and the point to find b. 3 = 1*2 + b b = 1 y = -x + 1 is thru (2,3) and perpendicular to y = –x – 3

OpenStudy (the_fizicx99):

Yes, x is the same as 1x so your new equation is; y-2=1x -3 what would you do now

OpenStudy (nurali):

Welcome to Openstudy @AlgebraOhNo

OpenStudy (anonymous):

You would add the 2 to the right side so you would isolate the y and then you would have y=x-1

OpenStudy (the_fizicx99):

Yup

OpenStudy (anonymous):

Thank you @Nurali

OpenStudy (the_fizicx99):

I would use 1x but its fine i guess

OpenStudy (nurali):

Anytime.

OpenStudy (anonymous):

Okay thank you!

OpenStudy (the_fizicx99):

This is 9th grade math, right? I think i remember doing this :o

OpenStudy (anonymous):

Im in 10th grade but yeah its Algebra 2

OpenStudy (the_fizicx99):

oh