If A (2, 2), B (−4, −4) and C (5, −8) are the vertices

Solution:

We have a triangle $\triangle \mathrm{ABC}$ in which the co-ordinates of the vertices are $\mathrm{A}(2,2) \mathrm{B}(-4,-4)$ and $\mathrm{C}(5,-8)$.

In general to find the mid-point $\mathrm{P}(x, y)$ of two points $\mathrm{A}\left(x_{1}, y_{1}\right)$ and $\mathrm{B}\left(x_{2}, y_{2}\right)$ we use section formula as,

$\mathrm{P}(x, y)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$

Therefore mid-point D of side AB can be written as,

$\mathrm{D}(x, y)=\left(\frac{2-4}{2}, \frac{2-4}{2}\right)$

Now equate the individual terms to get,

$x=-1$

$y=-1$

So co-ordinates of D is (−1,−1)

So the length of median from C to the side AB,

$\mathrm{CD}=\sqrt{(5+1)^{2}+(-8+1)^{2}}$

$=\sqrt{36+49}$

$=\sqrt{85}$

So the answer is (c)