Question

# For ∆ABC, whose vertices are A(4, −6), B(3, −2) and C(5, 2), verify that a median of the triangle divides it into two triangles of equal areas.

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Solution

## To prove : Area of triangle ABD = Area of triangle ACD Given ∆ABC, whose vertices are A(4, −6), B(3, −2) and C(5, 2). Join AD. Let AD be the median that divides the triangle into two triangles. So, D is the midpoint of BC. Therefore, the coordinates of D are $D\left(\frac{3+5}{2},\frac{-2+2}{2}\right)=D\left(\frac{8}{2},\frac{0}{2}\right)=D\left(4,0\right)$ Now, $\mathrm{Area}\mathrm{of}\mathrm{triangle}ABD=\frac{1}{2}\left\{{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\right\}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[4\left(-2-0\right)+3\left\{0-\left(-6\right)\right\}+4\left\{-6-\left(-2\right)\right\}\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left\{4\left(-2\right)+3\left(6\right)+4\left(-4\right)\right\}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left(-8+18-16\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left(-6\right)\phantom{\rule{0ex}{0ex}}=3\mathrm{sq}.\mathrm{units}.\left[\mathrm{Area}\mathrm{cannot}\mathrm{be}-\mathrm{ve}\right]$ and $\mathrm{Area}\mathrm{of}\mathrm{triangle}ACD=\frac{1}{2}\left\{{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\right\}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left\{4\left(2-0\right)+5\left(0-\left(-6\right)\right)+4\left(-6-2\right)\right\}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left\{4\left(2\right)+5\left(6\right)+4\left(-8\right)\right\}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left(8+30-32\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left(6\right)\phantom{\rule{0ex}{0ex}}=3\mathrm{sq}.\mathrm{units}.$ Therefore, area of triangle ABD = Area of triangle ACD Hence, the median of ∆ABC divides it into two triangles of equal areas.

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