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

The origin and the points where the line $L_1$ intersects the $x$-axis and $y$-axis are vertices of a right-angled triangle $T$ whose area is $8$. Also,the line $L_1$ is perpendicular to the line $L_2: 4x - y = 3$. Then,the perimeter of triangle $T$ is:

The line $x + y = p$ meets the $x$ and $y$ axes at $A$ and $B$ respectively. $A$ triangle $APQ$ is inscribed in the triangle $OAB$,where $O$ is the origin,with a right angle at $Q$. $P$ and $Q$ lie on $OB$ and $AB$ respectively. If the area of triangle $APQ$ is $3/8$ of the area of triangle $OAB$,then $\frac{AQ}{BQ}$ is equal to:

The line $3x + 2y = 24$ meets the $y$-axis at $A$ and the $x$-axis at $B$. The perpendicular bisector of $AB$ meets the line through $(0, -1)$ parallel to the $x$-axis at $C$. The area of the triangle $ABC$ is ............... $sq. \, units$.

If a line $ax + 2y = k$ forms a triangle of area $3$ sq. units with the coordinate axes and is perpendicular to the line $2x - 3y + 7 = 0$,then the product of all the possible values of $k$ is

$A$ 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