Distance to a line, without a formula

Squared distance. $y=\dfrac{10-3x}4$, so
$$D(x)=x^2+\frac{(10-3x)^2}{16}=\frac{25x^2-60x+100}{16}.$$
Minimising $D$ (the squared distance) is equivalent to minimising the distance, since the square root is increasing.

Minimum. $D'(x)=\dfrac{50x-60}{16}=0\iff x=\dfrac65$, hence $y=\dfrac{10-18/5}4=\dfrac85$. $D$ is a convex parabola, so this is the global minimum, and $D\left(\frac65\right)=\frac{36+64}{25}=4$.

Answer. The closest point is $\left(\frac65,\frac85\right)$ and the distance is $\sqrt4=2$.

Check. The formula $\dfrac{|3\cdot0+4\cdot0-10|}{\sqrt{3^2+4^2}}=\frac{10}5=2$ agrees. Also, the vector $\left(\frac65,\frac85\right)$ is proportional to the normal vector $(3,4)$ of the line, as it must be.