Points A (-10,-6) and B (6,2) are the endpoints of AB. What are the coordinates of point C on AB such that AC is 3/4 the length of AB? a. (0,-1) b. (2,0) c. (-2,-2) d. (4,1)

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Points A (-10,-6) and B (6,2) are the endpoints of AB. What are the coordinates of point C on AB such that AC is 3/4 the length of AB? a. (0,-1) b. (2,0) c. (-2,-2) d. (4,1)

Geometry
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Are you fond to vectors?
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No
Actually yes I am.
Okay, so I'll show you the vector way. We can express the vector whose tail is located on "A" and head on "B" like this: \[\vec a ((6-(-10)), (2-(-6))) \rightarrow \vec a (16,8)\] \(\vec a\) is a vector that goes from A to B, and we want to find a "C" on AB such that AC=3/4 the length of AB. This means a 3/4 part of the vector, this creates a new vector \(\vec w\) that is \(\frac{ 3 }{ 4 } \vec a\), but we can do it by expressing the vector on it's referential form: \[\vec a = 16 \vec i + 8 \vec j \rightarrow \frac{ 3 }{ 4 } \vec a = (\frac{ 3 }{ 4 } 16) \vec i + (\frac{ 3 }{ 4 }8) \vec j \] \[\frac{ 3 }{ 4 } \vec a = \vec w \] \[\vec w = (3.4) \vec i + (3.2)\vec j\] \[\vec w = 12 \vec i + 6 \vec j\] This implies that the new coordinates lie on \[\vec w (12,6)\] And this implies, regressing the operation of the vector, since the tail still lies on A, but the head is now on the point C: \[x_c-(-10)=12\] \[y_c - (-6)=6\]
Did it kick everyone offline?
No, just some changes in the website.
So it would be D?
no, solve the first equation for xc and the other for yc, that'll give you the coordinates of the point C
So B
correct.
When I said I was fond of vectors I ment in science...
Gave you a medal... Thanks so much.
Oh.. Well, if I were to say the method wihout them, I would suggest you find the distance between those two points. Then multiply it by 3/4, and then the process is completely analogous.

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