Question

what is the equation in point-slope form of the line passing through (0,6) and (1,3)

Answer 1

\(\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
&({\color{red}{ 0}}\quad ,&{\color{blue}{ 6}})\quad &({\color{red}{ 1}}\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}}}
\\ \quad \\

y-y_1={\color{green} m}(x-x_1)\Leftarrow\textit{point-slope form, plug in your values}\)

Answer 2

(y-3)=3(x+6)

Answer 3

am i right?

Answer 4

ahemm what did you get for the slope?

Answer 5

-3 and 6

Answer 6

hmm 6? well, -3 is right.. yes \(\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
&({\color{red}{ 0}}\quad ,&{\color{blue}{ 6}})\quad &({\color{red}{ 1}}\quad ,&{\color{blue}{ 3}})
\end{array}
\\\quad \\

slope = {\color{green}{ m}}= -3
\\ \quad \\

y-{\color{blue}{ 6}}=\color{green}{ -3}(x-{\color{red}{ 0}})\)

Answer 7

or y-6=-3x

Answer 8

so i was right? @jdoe0001

Answer 9

well... you had -> (y-3)=3(x+6) <-
the slope is correct, is -3

the (y-3)=3(x+6)
^ ^
indicates a point at (-6, 3) which you don't have

Answer 10

oh so it would be (y-3)=-3(x-1)

Answer 11

@jdoe0001

Answer 12

\(\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
&({\color{red}{ \large \bf 0}}\quad ,&{\color{blue}{\large \bf 6}})\quad &({\color{red}{ 1}}\quad ,&{\color{blue}{ 3}})
\end{array}
\\\quad \\

slope = {\color{green}{ m}}= -3
\\ \quad \\

\large y-{\color{blue}{ 6}}=\color{green}{ -3}(x-{\color{red}{ 0}})\)

Answer 13

you can really just use either... so


if we use the 2nd point you'd end up with -> (y-3)=-3(x-1) <-- yes
and that's valid too

Answer 14

since both equations will simplify to the same

Answer 15

thank you

Answer 16

yw