2. anonymous

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

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

4. anonymous

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

5. anonymous

plug in $$x_1\ and\ y_1$$ and the found slope

6. anonymous

so.. find the slope first

. . . and how do i do that

8. anonymous

well... see above

9. anonymous

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

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

11. anonymous

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

12. anonymous

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

and then i just divide them?

14. anonymous

or leave them as fraction, yes

|dw:1442534647425:dw| like this

16. anonymous

hmmm one sec

17. anonymous

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

3?

19. anonymous

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?

21. anonymous

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

22. anonymous

x-(-2) => x+2 so... bear in mind that

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

24. anonymous

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?

26. anonymous

nothing, be happy, eat ice-cream

oh..im done ?

28. anonymous

yeap

thank you so much :)

30. anonymous

yw