Straight Lines and Pair of Straight Lines

To find point $A$, we need to solve the equations:$\;3 x+y-4=0$$\;x-y=0$The second equation gives us $y=x$. Substitute $y=x$ into the first equation:$\;3 x+x-4=0$$\;4 x-4=0$$\;4 x=4$$\;x=1$Substituting $x=1$ into $y=x$, we get $y=1$.Thus, the coordinates of $A$ are $(1,1)$.Next, consider the line $x-3 y+5=0$ which has a slope of $\;\frac{1}{3}$.Let $m$ be the slope of the required line. Since the line makes an angle of $45^{\circ}$ with the given line and has a negative slope, we use the angle formula:$\tan 45^{\circ}=\left|\frac{\;\frac{1}{3}-m}{1+\;\frac{1}{3}\cdot m} \right|$Since $\tan 45^{\circ}=1$, we have:$1=\left|\frac{\;\frac{1-3m}{3}}{\frac{3+m}{3}} \right|$This simplifies to:$|3+m|=|1-3 m|$Solving $3+m= \pm (1-3 m)$ gives two possibilities:$\;3+m=1-3 m$$\;4 m=-2 \;\; \Rightarrow \;\; m=-\;\frac{1}{2}$$\;3+m=-1+3 m$$\;-2 m=-4 \;\; \Rightarrow \;\; m=2$Since the slope must be negative, we choose $m=-\;\frac{1}{2}$.The equation of the line using point-slope form is:$y-1=-\;\frac{1}{2}(x-1)$Simplifying:$\;2(y-1)=-(x-1)$$\;2 y-2=-x+1$$\;x+2 y=3$Thus, the required equation of the line is:$x+2 y=3$