Chapter 5 - Inner Product Spaces - 5.2 Orthogonal Sets of Vectors and Orthogonal Projections - Problems - Page 360: 24

$\frac{13}{\sqrt 2}$

We are given: $x-y=3$ Rewrite as: $y=x-3$ From exercise 21, we have: $d(P,L)=\frac{|y_0-x_0m-b|}{\sqrt 1+m^2}$ Since $P(-6,4)$ and $y=x-3$, hence: $$d(P,L)=\frac{|4-(-6).1-(-3)|}{\sqrt 1+1^2}=\frac{|4+6+3|}{\sqrt 2}=\frac{13}{\sqrt 2}$$