Question

Which is the equation in slope-intercept form of the line that contains points A and C?

y=-x-6
y+4=-1(x+2)
x+y=-6
x+-y-6

I got A

Answer 1

\(\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
A&({\color{red}{ -4}}\quad ,&{\color{blue}{ -2}})\quad C&({\color{red}{ -2}}\quad ,&{\color{blue}{ -4}})
\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-{\color{blue}{ y_1}}={\color{green}{ m}}(x-{\color{red}{ x_1}})\qquad \textit{plug in the values and solve for "y"}\\
\qquad \uparrow\\
\textit{point-slope form}\)

Answer 2

i must be doin it wrong cuz i got A @jdoe0001

Answer 3

hmmm

Answer 4

\(\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
A&({\color{red}{ -4}}\quad ,&{\color{blue}{ -2}})\quad C&({\color{red}{ -2}}\quad ,&{\color{blue}{ -4}})
\end{array}
\\\quad \\

slope = {\color{green}{ m}}= \cfrac{rise}{run} \implies \cfrac{{\color{blue}{ -4}}-{\color{blue}{ (-2)}}}{{\color{red}{ -2}}-{\color{red}{ (-4)}}}\to \cfrac{-4+2}{-2+4}\to \cancel{ \cfrac{-2}{+2} }\to -1
\\ \quad \\

y-{\color{blue}{ (-2)}}={\color{green}{ -1}}(x-{\color{red}{ (-4)}})\implies y+2=-(x+4)
\\ \quad \\
y+2=-x-4\implies y=-x-4-2\implies y=-x-6\)

Answer 5

i was right then aha

Answer 6

much mahalos @jdoe0001

Answer 7

yeap