adajiamcneal
  • adajiamcneal
please help
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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adajiamcneal
  • adajiamcneal
1 Attachment
jdoe0001
  • jdoe0001
\(\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({\color{red}{ -2}}\quad ,&{\color{blue}{ 15}})\quad % (c,d) &({\color{red}{ 9}}\quad ,&{\color{blue}{ -18}}) \end{array} \\\quad \\ % slope = m slope = {\color{green}{ m}}= \cfrac{rise}{run} \implies \cfrac{{\color{blue}{ y_2}}-{\color{blue}{ y_1}}}{{\color{red}{ x_2}}-{\color{red}{ x_1}}} \\ \quad \\ % point-slope intercept y-{\color{blue}{ y_1}}={\color{green}{ m}}(x-{\color{red}{ x_1}})\qquad \textit{plug in the values and solve for "y"}\\ \qquad \uparrow\\ \textit{point-slope form}\)
adajiamcneal
  • adajiamcneal
. . . . so i plug in the x and ys?

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jdoe0001
  • jdoe0001
actually, you don't need to solve for "y: since you're asked to leave it in point-slope form anyhow
jdoe0001
  • jdoe0001
plug in \(x_1\ and\ y_1\) and the found slope
jdoe0001
  • jdoe0001
so.. find the slope first
adajiamcneal
  • adajiamcneal
. . . and how do i do that
jdoe0001
  • jdoe0001
well... see above
jdoe0001
  • jdoe0001
\(\bf slope = {\color{green}{ m}}= \cfrac{rise}{run} \implies \cfrac{{\color{blue}{ y_2}}-{\color{blue}{ y_1}}}{{\color{red}{ x_2}}-{\color{red}{ x_1}}}\)
adajiamcneal
  • adajiamcneal
so i plug it in like i said or just x 1 and y 1
jdoe0001
  • jdoe0001
they do have a value, it's given in coordinates
jdoe0001
  • jdoe0001
\(\begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({\color{red}{ -2}}\quad ,&{\color{blue}{ 15}})\quad % (c,d) &({\color{red}{ 9}}\quad ,&{\color{blue}{ -18}}) \end{array}\)
adajiamcneal
  • adajiamcneal
and then i just divide them?
jdoe0001
  • jdoe0001
or leave them as fraction, yes
adajiamcneal
  • adajiamcneal
|dw:1442534647425:dw| like this
jdoe0001
  • jdoe0001
hmmm one sec
jdoe0001
  • jdoe0001
\(\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({\color{red}{ -2}}\quad ,&{\color{blue}{ 15}})\quad % (c,d) &({\color{red}{ 9}}\quad ,&{\color{blue}{ -18}}) \end{array} \\\quad \\ % slope = m slope = {\color{green}{ m}}= \cfrac{rise}{run} \implies \cfrac{{\color{blue}{ -18}}-{\color{blue}{ 15}}}{{\color{red}{ 9}}-{\color{red}{ (-2)}}}\implies \cfrac{-33}{9+2}\implies \cfrac{-\cancel{33}}{\cancel{11}}\implies ?\)
adajiamcneal
  • adajiamcneal
3?
jdoe0001
  • jdoe0001
well... don't miss the " - " in front of the 33 so is -3 so, that's the slope, so plug that in, along with \(x_1\ and\ y_1\)
adajiamcneal
  • adajiamcneal
wait so what do i do with the -3?
jdoe0001
  • jdoe0001
\(\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({\color{red}{ -2}}\quad ,&{\color{blue}{ 15}})\quad % (c,d) &({\color{red}{ 9}}\quad ,&{\color{blue}{ -18}}) \end{array} \\\quad \\ % slope = m slope = {\color{green}{ m}}= -3 \\ \quad \\ % point-slope intercept y-{\color{blue}{ 15}}={\color{green}{ -3}}(x-{\color{red}{ (-2)}})\\ \qquad \uparrow\\ \textit{point-slope form}\)
jdoe0001
  • jdoe0001
x-(-2) => x+2 so... bear in mind that
adajiamcneal
  • adajiamcneal
so do i add 15 to the 3? to get y
jdoe0001
  • jdoe0001
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 :)
adajiamcneal
  • adajiamcneal
so then what do i do?
jdoe0001
  • jdoe0001
nothing, be happy, eat ice-cream
adajiamcneal
  • adajiamcneal
oh..im done ?
jdoe0001
  • jdoe0001
yeap
adajiamcneal
  • adajiamcneal
thank you so much :)
jdoe0001
  • jdoe0001
yw

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