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
help
Did you need a second opinion? XD
yeah
lol
I was thinking A
what's the slope of 3x + y = 7 ?
3 I think?
can you put the equation in y = mx + b form ?
well... can you solve that for "y"?
y = 3x + 7 Is y = mx + b form
but i don't knwo what to do neext
well.... it should have been -3x... you'd subtract -3x from both sides
http://www.algebra-class.com/image-files/slope-formula-1.gif <--- this tells you what the slope of that is
opps and oh
so the slope is 3 ?
\(\bf\cancel{-3x+3x} + y = -3x+7\implies y=-3x+7\)
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)
okay
\(\bf \begin{array}{lllll} &x_1&y_1\\ &(6\quad ,&-1) \end{array} \\\quad \\ slope = m= \cfrac{1}{3} \\ \quad \\ y-y_1=m(x-x_1)\qquad \textit{plug in your values and solve for "y"}\)
so it would be either answer A or B
once you plug in the values in the "point-slope form", solving for "y" will give you the equation
y - 1 = 1/3(x - 6)
then i do the distributive property?
on the right side
well... yes however.... \(\bf \begin{array}{lllll} &x_1&y_1\\ &(6\quad ,&-1) \end{array} \\\quad \\ slope = m= \cfrac{1}{3} \\ \quad \\ y-y_1=m(x-x_1)\implies y-(-1)=\cfrac{1}{3}(x-6)\implies y+1=\cfrac{1}{3}(x-6)\)
The answer is y=(1/3)x-3
\(\bf y+1=\cfrac{1}{3}(x-6)\implies y+1=\cfrac{1}{3}x-\cfrac{6}{3}\implies y+1=\cfrac{1}{3}x-2\\ \quad \\ y\cancel{+1-1}=\cfrac{1}{3}x-2-1\)
graph the original equation in y=mx+b form
then graph the answers to see which one is perpendicular
did you see the answer choices at the top and sorry my laptop froze up
hello?
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That was much easier and so its A
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yup :)
thanks so much
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