Answer :
[tex]The\ distance\ between\ P(x_P;\ y_P)\ and\ the\ line\ Ax+By+C=0:\\\\d=\frac{|Ax_P+By_P+C|}{\sqrt{A^2+B^2}}\\===================================\\P(2;\ 5);\ 2x-y+7=0\\\\x_P=2;\ y_P=5;\ A=2;\ B=-1;\ C=7\\\\d=\frac{|2\times2-1\times5+7|}{\sqrt{2^2+(-1)^2}}=\frac{|4-5+7|}{\sqrt{4+1}}=\frac{|6|}{\sqrt5}\cdot\frac{\sqrt5}{\sqrt5}=\frac{6\sqrt5}{5}[/tex]
