[tex] \bf (\stackrel{x_1}{5}~,~\stackrel{y_1}{3})\qquad
(\stackrel{x_2}{2}~,~\stackrel{y_2}{-1})
\\\\\\
slope = m\implies
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-1-3}{2-5}\implies \cfrac{-4}{-3}\implies \cfrac{4}{3}
\\\\\\
\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-3=\cfrac{4}{3}(x-5)\implies y-3=\cfrac{4}{3}x-\cfrac{20}{3}
\\\\\\
y=\cfrac{4}{3}x-\cfrac{20}{3}+3\implies y=\cfrac{4}{3}x-\cfrac{11}{3} [/tex]
First, we must determine the slope:
Slope = m = (Y2 -Y1) ÷ (X2 -X1)
Slope = (-1, -3) / (2 -5)
Slope = -4 / -3 = 4 / 3 = 1.33333333...
Now, we have to fill in this equaton:
y = mx + b we know the slope "m"
y = 1.3333... x + b now we'll take a point (5, 3) and solve for "b"
b = y -mx b = 3 - 1.3333... *5
b = -3.6666666667
y = 1.3333... x -3.6666...
Source
1728.com/distance.htm
