## Calculus 10th Edition

$(\frac{b}{3},\frac{c}{3})$
To find the point of intersection, we must find the equation of at least 2 of the medians. That can be found by calculating the midpoints of each side and then the slope of each median. 1. Equation of median passing through the top corner: $midpoint=(\frac{-a+a}{2},\frac{0+0}{2})=(0,0)$ $slope=\frac{c-0}{b-0}=\frac{c}{b}$ Equation: $y-0=\frac{c}{b} (x-0)$ $y=\frac{cx}{b}$ 2. Equation of median passing through the bottom-left corner: $midpoint = (\frac{a+b}{2},\frac{c+0}{2})= (\frac{a+b}{2},\frac{c}{2})$ $slope=\frac{\frac{c}{2}-0}{\frac{a+b}{2}+a}=\frac{\frac{c}{2}}{\frac{3a+b}{2}}=\frac{c}{3a+b}$ Equation: $y-\frac{c}{2}=\frac{c}{3a+b}(x-\frac{a+b}{2})$ Now we can substitute $y = \frac{cx}{b}$ to solve for x: $\frac{cx}{b}-\frac{c}{2}=\frac{c}{3a+b}(x-\frac{a+b}{2})$ $\frac{2cx}{2b}-\frac{bc}{2b}=\frac{c}{3a+b} (\frac{2x}{2}-\frac{a+b}{2})$ $\frac{2cx-bc}{2b}=\frac{c}{3a+b} (\frac{2x-a-b}{2})$ $\frac{2x-b}{b}=\frac{2x-a-b}{3a+b}$ [multiply both sides by $\frac{2}{c}$] $6ax-3ab+2xb-b^2=2xb-ab-b^2$ [cross multiply] $6ax=2ab$ $x=\frac{b}{3}$ Since we have solved x, it is time to go back and solve y: $y=\frac{cx}{b}$ $y=\frac{c}{b}\frac{b}{3}$ $y=\frac{c}{3}$ The intersection is $(\frac{b}{3},\frac{c}{3})$