The equation of the straight line perpendicular to $5x - 2y = 7$ and passing through the point of intersection of the lines $2x + 3y = 1$ and $3x + 4y = 6$ is

$A$ straight line passing through the origin $O$ meets the parallel lines $4x + 2y = 9$ and $2x + y + 6 = 0$ at the points $P$ and $Q$ respectively. Then the point $O$ divides the line segment $PQ$ in the ratio:

Given the family of lines,$a(2x + y + 4) + b(x - 2y - 3) = 0$. Among the lines of the family,the number of lines situated at a distance of $\sqrt{10}$ from the point $M(2, -3)$ is:

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:

If $16a^2 - 40ab + 25b^2 - c^2 = 0$,then the line $ax + by + c = 0$ passes through which points?