A straight line through Q(2, 3) makes on angles $\frac{3\mathrm{\pi }}{4}$ with the negative direction of the x-axis. If the straight line intersects the line x + y - 7 = 0 at P, find the distance PQ.

Slope of the line is m
where $\mathrm{m}=\tan \left(\mathrm{\pi }-\frac{3\mathrm{\pi }}{4}\right)=\tan \frac{\mathrm{\pi }}{4}=1$
Equation of the line passing through Q(2,3) and $\text{ slope }1\text{ is }\mathrm{y}-3=1(\mathrm{x}-2)$
$\mathrm{x}-\mathrm{y}+1=0\dots \left(1\right)$
$\text{ Given line }\mathrm{x}+\mathrm{y}-7=0 ...\left(2\right)$
Point of intersection of (1) and (2) is (3, 4)
$\text{ Distance }\stackrel{ }{\mathrm{PQ}}=\sqrt{(3-2{)}^2+(4-3{)}^2}=\sqrt{1+1}=\sqrt{2}$