please help

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

actually, you don't need to solve for "y: since you're asked to leave it in point-slope form anyhow
plug in \(x_1\ and\ y_1\) and the found slope
so.. find the slope first
. . . and how do i do that
well... see above
\(\bf slope = {\color{green}{ m}}= \cfrac{rise}{run} \implies \cfrac{{\color{blue}{ y_2}}-{\color{blue}{ y_1}}}{{\color{red}{ x_2}}-{\color{red}{ x_1}}}\)
so i plug it in like i said or just x 1 and y 1
they do have a value, it's given in coordinates
\(\begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({\color{red}{ -2}}\quad ,&{\color{blue}{ 15}})\quad % (c,d) &({\color{red}{ 9}}\quad ,&{\color{blue}{ -18}}) \end{array}\)
and then i just divide them?
or leave them as fraction, yes
|dw:1442534647425:dw| like this
hmmm one sec
\(\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({\color{red}{ -2}}\quad ,&{\color{blue}{ 15}})\quad % (c,d) &({\color{red}{ 9}}\quad ,&{\color{blue}{ -18}}) \end{array} \\\quad \\ % slope = m slope = {\color{green}{ m}}= \cfrac{rise}{run} \implies \cfrac{{\color{blue}{ -18}}-{\color{blue}{ 15}}}{{\color{red}{ 9}}-{\color{red}{ (-2)}}}\implies \cfrac{-33}{9+2}\implies \cfrac{-\cancel{33}}{\cancel{11}}\implies ?\)
3?
well... don't miss the " - " in front of the 33 so is -3 so, that's the slope, so plug that in, along with \(x_1\ and\ y_1\)
wait so what do i do with the -3?
\(\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({\color{red}{ -2}}\quad ,&{\color{blue}{ 15}})\quad % (c,d) &({\color{red}{ 9}}\quad ,&{\color{blue}{ -18}}) \end{array} \\\quad \\ % slope = m slope = {\color{green}{ m}}= -3 \\ \quad \\ % point-slope intercept y-{\color{blue}{ 15}}={\color{green}{ -3}}(x-{\color{red}{ (-2)}})\\ \qquad \uparrow\\ \textit{point-slope form}\)
x-(-2) => x+2 so... bear in mind that
so do i add 15 to the 3? to get y
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 :)
so then what do i do?
nothing, be happy, eat ice-cream
oh..im done ?
yeap
thank you so much :)
yw

Not the answer you are looking for?

Search for more explanations.

Ask your own question