Choose the point-slope form of the equation below that represents the line that passes through the points (-3, 2) and (2, 1). y + 3 = -5(x - 2) y - 2 = -5(x + 3) y + 3 = -one fifth(x - 2) y - 2 = -one fifth(x + 3)
\(\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ &({\color{red}{ -3}}\quad ,&{\color{blue}{ 2}})\quad &({\color{red}{ 2}}\quad ,&{\color{blue}{ 1}}) \end{array} \\\quad \\ slope = {\color{green}{ m}}= \cfrac{rise}{run} \implies \cfrac{{\color{blue}{ y_2}}-{\color{blue}{ y_1}}}{{\color{red}{ x_2}}-{\color{red}{ x_1}}} \\ \quad \\ y-{\color{blue}{ y_1}}={\color{green}{ m}}(x-{\color{red}{ x_1}})\qquad \textit{plug in the values}\\ \qquad \uparrow\\ \textit{point-slope form}\)
y-4=M(x-(-3))?
yes.... .m = slope just get the slope and "stick it" in there :)
well.... hmm when using the coordinates you'd want to use the coordinates from the same point
1-2/2-(-3)
-1/5?
so if you use -3, you'd need to use 2 if you use 2, you'd need to use 1 or if you use the "x" of the 1st point, you'd have to use the "y" for the same 1st one
yeap -1/5 = m= slope \(\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ &({\color{red}{ -3}}\quad ,&{\color{blue}{ 2}})\quad &({\color{red}{ 2}}\quad ,&{\color{blue}{ 1}}) \end{array} \\\quad \\ slope = {\color{green}{ m}}= \cfrac{rise}{run} \implies \cfrac{{\color{blue}{1}}-{\color{blue}{ 2}}}{{\color{red}{ 2}}-{\color{red}{ (-3)}}}\implies \cfrac{-1}{2+3}\implies -\cfrac{1}{5} \\ \quad \\ y-{\color{blue}{ 2}}={\color{green}{ -\cfrac{1}{5}}}(x-{\color{red}{ (-3)}})\qquad or \\ \quad \\ y-{\color{blue}{ 1}}={\color{green}{ -\cfrac{1}{5}}}(x-{\color{red}{2}})\)
y-2=-1/5(x-(-3))
or y-2=-5(x+3)
yeap well \(\bf y-{\color{blue}{ 2}}={\color{green}{ -\cfrac{1}{5}}}(x-{\color{red}{ (-3)}})\implies y-{\color{blue}{ 2}}={\color{green}{ -\cfrac{1}{5}}}(x+{\color{red}{ 3}})\)
-1/5=-5
simplifies it
s###
no it doesnt
dur
hmm \(\bf -\cfrac{5}{1}\iff -5\qquad \qquad -\cfrac{1}{5}\iff -\cfrac{1}{5}\)
welp, Thanks alot man
g2g
yw
Join our real-time social learning platform and learn together with your friends!