What is the slope of the line that passes through (-3, 3) and (-3, 6)? A) -1/2 B) 0 C) 1/2 D) Undefined
\[\LARGE \frac{y_2-y_1}{x_2-x_1}=slope\]
so, \[\frac{ 6 }{ -3 } - \frac{ 3 }{ -3 }\] which equals \[\frac{ 3 }{ 0 }\]
i think this is undefined
why do you say that? @needhelpexplainplz
because it makes a vertical line
Oh, is it my work at the top correct though?
yes i believe so
A vertical line has "undefined slope" because all points on the line have the same x-coordinate.
you can close your question now
I know..but i can't do that whenever i still don't get the question. I don't want the answer just like you and @abb0t gave me. I want an explanation to why it is that.
@Deadly_Roses @RCCB
They already gave you an explantion: Slope = m = \(\sf \large \frac{\Delta y }{\Delta x} = \frac{y_2-y_1}{x_2-x_1}\) you subtract the second coordinate from the first. it is the change in coordinates.
@abb0t Don't even mind replying back to what i just said above, i could careless about what your peasant mind has to say to me so therefore leave, thank you have a nice day(:
@https.wild_ um excuse me i know your talking to him but you dont have a reason to be rude we shouldnt have given you the answer >:(
I'm disappointed that you fail to understand and grasp how this problems works, despite the explanations given to you by multiple users. I suppose you need to study your elementary mathematics.
I didn't wanted the answer..i wanted an explanation @needhelpexplainplz
i did give you the explaination i said that it was that answer because it was a veritical line!
No one gave you the answer here. Close the question, you're just trolling now.
how to find the slope.....(this is on a different problem, but just to show you how) To find the slope, we will need two points from the line. Pick two x's and solve for each corresponding y: If, say, x = 3, then y = ( 2/3 )(3) – 4 = 2 – 4 = –2. If, say, x = 9, then y = ( 2/3 )(9) – 4 = 6 – 4 = 2. (By the way, I picked the x-values to be multiples of three because of the fraction.
Thank you, that makes more sense to me. @Deadly_Roses
your welcome :)
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