The distance of the line 3x-y=0 from the point (4, 1) measured along a line making an angle 135⁰ with the x-axis is

$\begin{array}{l}Givenline3x-y=0--\left(1\right)\\ Letp(4,1)and\theta =135\\ m=\mathrm{Tan}\theta =\mathrm{Tan}{135}^0=-1\\ Equationoflinepas\sin gthroughP(4,1)is\\ y-1=-1(x-4)\\ \Rightarrow x+y-5=0--\left(2\right)\\ \left(1\right)+\left(2\right)\Rightarrow 4x=5\Rightarrow x=\frac{5}{4}\\ from\left(1\right)y=3\left(\frac{5}{4}\right)=\frac{15}{4}\\ let\text{\hspace{0.17em}Q}(\frac{5}{4},\frac{15}{4})\\ \therefore PQ=\sqrt{{(\frac{5}{4}-4)}^2+{(\frac{15}{4}-1)}^2}=\sqrt{\frac{121}{16}+\frac{121}{16}}=11\sqrt{\frac{2}{16}}=\frac{11\sqrt{2}}{4}\end{array}$