$A$ is the point of intersection of the lines $3x + y - 4 = 0$ and $x - y = 0$. If a line having a negative slope makes an angle of $45^{\circ}$ with the line $x - 3y + 5 = 0$ and passes through $A$,then its equation is:

Let $P$ be the point of intersection of the lines $L_1 \equiv x-y-7=0$ and $L_2 \equiv x+y-5=0$. $A(x_1, y_1)$ and $B(x_2, y_2)$ are points on the lines $L_1=0$ and $L_2=0$ respectively such that $PA=3\sqrt{2}$,$PB=\sqrt{2}$,$x_1, y_1 \geq 0$,$x_2, y_2 \geq 0$. Then the angle made by the line segment $AB$ at the origin is

If $a, b, c$ are in $A.P.$,then the straight line $ax + by + c = 0$ will always pass through the point

$A$ straight line which makes equal intercepts on positive $X$ and $Y$ axes and which is at a distance $1$ unit from the origin intersects the straight line $y=2x+3+\sqrt{2}$ at $(x_0, y_0)$. Then $2x_0+y_0$ is equal to

The point $(4,1)$ undergoes the following transformations successively:
$(i)$ Reflection in the line $x-y=0$
(ii) Shifting through a distance of $2$ units along the positive $X$-axis
(iii) Projection on the $X$-axis
The coordinates of the point in its final position are

$A$ straight line is drawn through the point $A(1,2)$ such that its point of intersection with the straight line $x+y=4$ is at a distance $\frac{\sqrt{6}}{3}$ from the given point $A$. Find the angle which the line makes with the positive direction of $X$-axis.