[Support area. The area helps to solve the problem] Heron's formula

The distances from point $M$, lying inside triangle $A B C$, to its sides $A C$ and $B C$ are 2 and 4, respectively. Find the distance from point $M$ to line $A B$, if $A B=10, B C=17, A C=21$.
Connect the given point $M$ with the vertices of the triangle $A B C$ and consider the areas of the three resulting triangles.

## Solution

By Heron's formula

$$
S_{\triangle \mathrm{ABC}}=\sqrt{24 \cdot 3 \cdot 7 \cdot 14}=7 \cdot 3 \cdot 4=84
$$

Let $P$ and $Q$ be the projections of point $M$ onto sides $B C$ and $A C$, and $T$ onto side $A B$. Then

$$
\begin{gathered}
S_{\triangle \mathrm{ABC}}=S_{\Delta \mathrm{BMC}}+S_{\triangle \mathrm{AMC}}+S_{\Delta \mathrm{AMB}}= \\
=\frac{1}{2} B C \cdot M P+\frac{1}{2} A C \cdot M Q+\frac{1}{2} A B \cdot M T= \\
=\frac{1}{2} \cdot 17 \cdot 4+\frac{1}{2} \cdot 21 \cdot 2+\frac{1}{2} \cdot 10 \cdot M T=34+21+5 M T
\end{gathered}
$$

From this, we find that

$$
M T=\frac{S_{\triangle A B C}-34-21}{5}=\frac{84-34-21}{5}=\frac{29}{5}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_b9b3326e3456e69a696eg-20.jpg?height=406&width=594&top_left_y=1818&top_left_x=729)

## Answer

$\frac{29}{5}$.