Write the equation of a line passing through the points (3,-4) and (2,6) for fan and medal!
@mathstudent55
@mathmale
@Hero
@jdoe0001
@zzr0ck3r
Do you know how to find the slope given two points?
yep!
Okay, then do that first.
-2
is the slope
so what do i do next!
\(\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ &({\color{red}{ 3}}\quad ,&{\color{blue}{ -4}})\quad &({\color{red}{ 2}}\quad ,&{\color{blue}{ 6}}) \end{array} \\\quad \\ slope = {\color{green}{ m}}= \cfrac{rise}{run} \implies \cfrac{{\color{blue}{ y_2}}-{\color{blue}{ y_1}}}{{\color{red}{ x_2}}-{\color{red}{ x_1}}} \\ \quad \\ y-{\color{blue}{ y_1}}={\color{green}{ m}}(x-{\color{red}{ x_1}})\qquad \textit{plug in the values and solve for "y"}\\ \qquad \uparrow\\ \textit{point-slope form}\)
so y=-2x+b
No. The point-slope equation is y-y1=m(x-x1) Choose any of the points that you listed in the original equation. I will choose 3,-4. y+4=-2(x-3) y+4=-2x+6 -4 -4 y=-2x+2
Actually you did your slope wrong.
\[\frac{ y2-y1 }{ x2-x1 } = \frac{ 6+4 }{ 2-3 }=\frac{ 10 }{ -1 }=-10\] NOW we can use the original points and plug them into the point-slope equation. y+4=-10(x-3) y+4=-10x+30 -4 -4 y=-10x+26 Using this equation we can double check our work with the other point, 2,6. 6=-10(2)+26 6=-20+26 6=6 So yes, that is the proper equation.
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