Question

please help

Answer 1

\(\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({\color{red}{ -2}}\quad ,&{\color{blue}{ 15}})\quad
% (c,d)
&({\color{red}{ 9}}\quad ,&{\color{blue}{ -18}})
\end{array}
\\\quad \\
% slope = m
slope = {\color{green}{ m}}= \cfrac{rise}{run} \implies
\cfrac{{\color{blue}{ y_2}}-{\color{blue}{ y_1}}}{{\color{red}{ x_2}}-{\color{red}{ x_1}}}
\\ \quad \\
% point-slope intercept
y-{\color{blue}{ y_1}}={\color{green}{ m}}(x-{\color{red}{ x_1}})\qquad \textit{plug in the values and solve for "y"}\\
\qquad \uparrow\\
\textit{point-slope form}\)

Answer 2

. . . . so i plug in the x and ys?

Answer 3

actually, you don't need to solve for "y:
since you're asked to leave it in point-slope form anyhow

Answer 4

plug in \(x_1\ and\ y_1\) and the found slope

Answer 5

so.. find the slope first

Answer 6

. . . and how do i do that

Answer 7

well... see above

Answer 8

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

Answer 9

so i plug it in like i said or just x 1 and y 1

Answer 10

they do have a value, it's given in coordinates

Answer 11

\(\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({\color{red}{ -2}}\quad ,&{\color{blue}{ 15}})\quad
% (c,d)
&({\color{red}{ 9}}\quad ,&{\color{blue}{ -18}})
\end{array}\)

Answer 12

and then i just divide them?

Answer 13

or leave them as fraction, yes

Answer 14

like this

Answer 15

hmmm
one sec

Answer 16

\(\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({\color{red}{ -2}}\quad ,&{\color{blue}{ 15}})\quad
% (c,d)
&({\color{red}{ 9}}\quad ,&{\color{blue}{ -18}})
\end{array}
\\\quad \\
% slope = m
slope = {\color{green}{ m}}= \cfrac{rise}{run} \implies
\cfrac{{\color{blue}{ -18}}-{\color{blue}{ 15}}}{{\color{red}{ 9}}-{\color{red}{ (-2)}}}\implies \cfrac{-33}{9+2}\implies \cfrac{-\cancel{33}}{\cancel{11}}\implies ?\)

Answer 17

3?

Answer 18

well... don't miss the " - " in front of the 33
so is -3

so, that's the slope, so plug that in, along with \(x_1\ and\ y_1\)

Answer 19

wait so what do i do with the -3?

Answer 20

\(\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({\color{red}{ -2}}\quad ,&{\color{blue}{ 15}})\quad
% (c,d)
&({\color{red}{ 9}}\quad ,&{\color{blue}{ -18}})
\end{array}
\\\quad \\
% slope = m
slope = {\color{green}{ m}}= -3
\\ \quad \\
% point-slope intercept
y-{\color{blue}{ 15}}={\color{green}{ -3}}(x-{\color{red}{ (-2)}})\\
\qquad \uparrow\\
\textit{point-slope form}\)

Answer 21

x-(-2) => x+2

so... bear in mind that

Answer 22

so do i add 15 to the 3? to get y

Answer 23

well... if you need it in "slope-intercept" form, yes, you could solve for "y:, yes
BUT
you're asked to leave it in "point-slope" form :)

Answer 24

so then what do i do?

Answer 25

nothing, be happy, eat ice-cream

Answer 26

oh..im done ?

Answer 27

yeap

Answer 28

thank you so much :)

Answer 29

yw