Let the line $2x + 3y = 18$ intersect the $Y$-axis at $B$. Suppose $C(\neq B)$,with coordinates $(a, b)$,is a point on the line such that $PB = PC$,where $P = (10, 10)$. Then,$8a + 2b$ equals

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

The points $(1, 1)$,$(0, \sec^2 \theta)$,and $(\csc^2 \theta, 0)$ are collinear for:

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

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	$IV$. $x+2y-10=0$
	$V$. $5x+5y+3=0$

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