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Mathematics 18 Online
OpenStudy (anonymous):

please help...Write the equation of the line that is perpendicular to the line 3x + y = 7 and passes through the point (6, −1). y = 1/3x - 3 y = 1/3x + 17 y = - 3x - 3 y = -3x + 17

OpenStudy (anonymous):

help

OpenStudy (anonymous):

Did you need a second opinion? XD

OpenStudy (anonymous):

yeah

OpenStudy (anonymous):

lol

OpenStudy (anonymous):

I was thinking A

OpenStudy (jdoe0001):

what's the slope of 3x + y = 7 ?

OpenStudy (anonymous):

3 I think?

OpenStudy (anonymous):

can you put the equation in y = mx + b form ?

OpenStudy (jdoe0001):

well... can you solve that for "y"?

OpenStudy (anonymous):

y = 3x + 7 Is y = mx + b form

OpenStudy (anonymous):

but i don't knwo what to do neext

OpenStudy (jdoe0001):

well.... it should have been -3x... you'd subtract -3x from both sides

OpenStudy (jdoe0001):

http://www.algebra-class.com/image-files/slope-formula-1.gif <--- this tells you what the slope of that is

OpenStudy (anonymous):

opps and oh

OpenStudy (anonymous):

so the slope is 3 ?

OpenStudy (jdoe0001):

\(\bf\cancel{-3x+3x} + y = -3x+7\implies y=-3x+7\)

OpenStudy (jdoe0001):

so the slope is -3 a perpendicular line to that, will have a NEGATIVE RECIPROCAL slope what does that mean? well, NEGATIVE of -3? +3 RECIPROCAL of that? 1/3 so now we know that the line perpendicular to 3x + y = 7 has a slope of 1/3 and passes through (6, -1)

OpenStudy (anonymous):

okay

OpenStudy (jdoe0001):

\(\bf \begin{array}{lllll} &x_1&y_1\\ &(6\quad ,&-1) \end{array} \\\quad \\ slope = m= \cfrac{1}{3} \\ \quad \\ y-y_1=m(x-x_1)\qquad \textit{plug in your values and solve for "y"}\)

OpenStudy (anonymous):

so it would be either answer A or B

OpenStudy (jdoe0001):

once you plug in the values in the "point-slope form", solving for "y" will give you the equation

OpenStudy (anonymous):

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

OpenStudy (anonymous):

then i do the distributive property?

OpenStudy (anonymous):

on the right side

OpenStudy (jdoe0001):

well... yes however.... \(\bf \begin{array}{lllll} &x_1&y_1\\ &(6\quad ,&-1) \end{array} \\\quad \\ slope = m= \cfrac{1}{3} \\ \quad \\ y-y_1=m(x-x_1)\implies y-(-1)=\cfrac{1}{3}(x-6)\implies y+1=\cfrac{1}{3}(x-6)\)

OpenStudy (anonymous):

The answer is y=(1/3)x-3

OpenStudy (jdoe0001):

\(\bf y+1=\cfrac{1}{3}(x-6)\implies y+1=\cfrac{1}{3}x-\cfrac{6}{3}\implies y+1=\cfrac{1}{3}x-2\\ \quad \\ y\cancel{+1-1}=\cfrac{1}{3}x-2-1\)

OpenStudy (anonymous):

graph the original equation in y=mx+b form

OpenStudy (anonymous):

then graph the answers to see which one is perpendicular

OpenStudy (anonymous):

did you see the answer choices at the top and sorry my laptop froze up

OpenStudy (anonymous):

hello?

OpenStudy (anonymous):

|dw:1386279373944:dw|

OpenStudy (anonymous):

That was much easier and so its A

OpenStudy (anonymous):

|dw:1386279555195:dw|

OpenStudy (anonymous):

yup :)

OpenStudy (anonymous):

thanks so much

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