$A$ 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:

If a straight line $L$ perpendicular to the line $3x - 4y = 6$ forms a triangle of area $6$ square units with the coordinate axes,then the minimum perpendicular distance from the point $(1, 1)$ to the line $L$ 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

The line $2x + 3y = 12$ meets the $x$-axis at $A$ and $y$-axis at $B$. The line through $(5, 5)$ perpendicular to $AB$ meets the $x$-axis,$y$-axis,and the $AB$ at $C, D,$ and $E$ respectively. If $O$ is the origin of coordinates,then the area of $OCEB$ is

If $\alpha$ is the angle made by the perpendicular drawn from the origin to the line $3x - 4y + 5 = 0$ with the positive $X$-axis in the positive direction,and $ax + by = 1$ is the equation of a line passing through the point $(1, -1)$ with $\tan \alpha$ as its slope,then $a + ab + b =$

Match the following:

List-$I$	List-$II$
$A$. The equation of line passing through $(4,3)$ whose $X$-intercept is twice its $Y$-intercept	$I$. $x+y-2\sqrt{2}=0$
$B$. The equation of the line passing through the centroid and circumcentre of $\triangle ABC$ with vertices $A(1,1), B(3,3), C(6,-6)$	$II$. $7x+23y-8=0$
$C$. The equation of the line whose $X$-intercept is $(-3/5)$ and is perpendicular to $x-y+2=0$	$III$. $x+2y+\sqrt{2}=0$
$D$. The equation of the line whose distance from the origin is $2$ and the normal drawn from the origin makes an angle $45^{\circ}$ with the positive direction of $X$-axis	$IV$. $x+2y-10=0$
	$V$. $5x+5y+3=0$