Question

HELP! what is the magnitude of the vector with an initial point (2,1) and terminal point (-13,9)

Answer 1

\(\bf \textit{distance between 2 points, or vector magnitude}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
&({\color{red}{ 2}}\quad ,&{\color{blue}{ 1}})\quad &({\color{red}{ -13}}\quad ,&{\color{blue}{ 9}})
\end{array}\qquad
d = \sqrt{({\color{red}{ x_2}}-{\color{red}{ x_1}})^2 + ({\color{blue}{ y_2}}-{\color{blue}{ y_1}})^2}\)

Answer 2

is the answer 7 ?

Answer 3

is that a guess? becuase its not close at all

Answer 4

no i worked out the problem he gave me

Answer 5

@jdoe0001 can you help me some more

Answer 6

well... anything confusing there? looks pretty straight forward

Answer 7

is it 7 or am i wrong

Answer 8

well what did you get for \(\bf (x_2-x_1)^2\) ?

Answer 9

one problem i have with the formula is i simply put the numbers in the wrong places ....

i would rather step it out to avoid the error

Answer 10

(-15)^2

Answer 11

yeap that'd 225, then get the other and sum them up :)

Answer 12

\(\bf \bf \textit{distance between 2 points, or vector magnitude}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
&({\color{red}{ 2}}\quad ,&{\color{blue}{ 1}})\quad &({\color{red}{ -13}}\quad ,&{\color{blue}{ 9}})
\end{array}\qquad
d = \sqrt{({\color{red}{ -13}}-{\color{red}{ 2}})^2 + ({\color{blue}{ 9}}-{\color{blue}{ 1}})^2}
\\ \quad \\
d=\sqrt{(-15)^2+(8)^2}\implies d=\sqrt{225+64}\)

Answer 13

so its 17

Answer 14

yeap

Answer 15

can you help me with another problem

Answer 16

easier if you post anew, if we dunno then someone else may know, we can also revise each other, also I'd be dashing in a few mins

Answer 17

ok