$A$ is the point of intersection of the lines $3 x+y-4=0$ and $x-y=0$. If a line having negative slope makes…
$A$ is the point of intersection of the lines $3 x+y-4=0$ and $x-y=0$. If a line having negative slope makes an angle of $45^{\circ}$ with the line $x-3 y+5=0$ and passes through A then its equation is
$x+y=2$
$x+2 y=3$
$4 x+3 y=7$
$x+3 y=4$
Solution
Since, intersection point of the lines $3 x+y-4=0$ $x-y=0$ is $\mathrm{A}(1,1)$
Now, slope of line $x-3 y+5=0$ is $\frac{1}{3}$
and slope of required line (m)
$=\left|\frac{\tan 45^{\circ}-\frac{1}{3}}{1+\frac{\tan 45^{\circ}}{3}}\right|=\left|\frac{1-\frac{1}{3}}{1+\frac{1}{3}}\right|=\frac{-2}{4}=\frac{-1}{2}$
So, equation of required line is
$y-1=\frac{-1}{2}(x-1) \Rightarrow y=\frac{-x}{2}+\frac{1}{2}+1 \Rightarrow 2 y+x=3$