[Coordinate method on the plane]

Find the distance between point $A(1,7)$ and the intersection point of the lines $x-y-1=0$ and $x+3 y-12=0$.

#
Solving the system, we find the point of intersection $B\left({ }^{15} / 4,{ }^{11} / 4\right)$ of the given lines. Using the formula for the distance between two points

$A B^{2}=\left({ }^{15} / 4-1\right)^{2}+\left({ }^{11 / 4}-1\right)^{2}=(11 / 4)^{2}+\left({ }^{17 / 4}\right)^{2}=1 / 16+(121+289)=410 / 16$.

## Answer

$\frac{\sqrt{410}}{4}$.