Question

Which point is collinear with points A and C?

A.
(0, 0)

B.
(1, 0)

C.
(0, –1)

D.
(–2, 2)


http://static.k12.com/calms_media/media/202500_203000/202592/1/8a992e88dbbabd6dcaf1dd91a8c102cfd382b597/28851_000515-23_ChoiceD.jpg

Answer 1

http://www.mathsisfun.com/definitions/collinear.html

Answer 2

that just lets me move the letters.....

Answer 3

have you covered slopes of lines yet?

Answer 4

nope

Answer 5

so... how are you supposed to find which of the given choices are collinear anyway?

Answer 6

i was absent and i dont understand this

Answer 7

well.... you find the slope between points A and C
and any other point " P "that is collinear to A and C, will also have the same slope

so when testing for say A and P, should give the same slope as A and C
and when testing for C and P, will also give the same slope

but you may want to cover the slopes sections first -> http://www.mathsisfun.com/geometry/slope.html

Answer 8

@Loser66

Answer 9

notice that point A is at (3,-4)
and point C is at (-4, 3)
thus the slope of points A and C would be \(\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
&({\color{red}{ 3}}\quad ,&{\color{blue}{ -4}})\quad &({\color{red}{ -4}}\quad ,&{\color{blue}{ 3}})
\end{array}
\\\quad \\

slope = {\color{green}{ m}}= \cfrac{rise}{run} \implies \cfrac{{\color{blue}{ y_2}}-{\color{blue}{ y_1}}}{{\color{red}{ x_2}}-{\color{red}{ x_1}}}\)

once you find their slope
do the same testing with the given choices

Answer 10

so @jdoe0001 it would be \[\frac{ 3-(-4) }{ (-4)-3 }\]

Answer 11

yeap that makes it \(\bf \cfrac{ 3-(-4) }{ (-4)-3 }\implies \cfrac{3+4}{-4-3}\implies \cfrac{\cancel{ 7 }}{-\cancel{ 7 }}\implies -1\leftarrow AC\ slope\)

Answer 12

-1

Answer 13

so... we know that the slope of AC is -1
so let us test say the 1st choice

say picking either A or C and the given point so (0,0) \(\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
A&({\color{red}{ 3}}\quad ,&{\color{blue}{ -4}})\quad &({\color{red}{ 0}}\quad ,&{\color{blue}{ 0}})
\end{array}
\\\quad \\

slope = {\color{green}{ m}}= \cfrac{rise}{run} \implies \cfrac{{\color{blue}{ 0}}-{\color{blue}{ 3}}}{{\color{red}{ 0}}-{\color{red}{ (-4)}}}\implies \cfrac{-3}{4}\ne -1\)

so the slope of A and point (0,0) is not -1...thus they're NOT collinear

Answer 14

so it would be A (0,0)

Answer 15

hmm hold the mayo.... messed up a bit...lemme fix that quick

Answer 16

okay :)

Answer 17

\(\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
A&({\color{red}{ 3}}\quad ,&{\color{blue}{ -4}})\quad &({\color{red}{ 0}}\quad ,&{\color{blue}{ 0}})
\end{array}
\\\quad \\

slope = {\color{green}{ m}}= \cfrac{rise}{run} \implies \cfrac{{\color{blue}{ 0}}-{\color{blue}{ (-4)}}}{{\color{red}{ 0}}-{\color{red}{ (3)}}}\implies \cfrac{4}{-3}\ne -1\)

so anyhow.. still no dice on that one