The equation of the line passing through the intersection of $3x - 4y + 1 = 0$ and $5x + y - 1 = 0$ which cuts off equal intercepts on the axes is given by

Among the statements:
$(S1)$ : If $A(5, -1)$ and $B(-2, 3)$ are two vertices of a triangle, whose orthocentre is $(0, 0)$, then its third vertex is $(-4, -7)$ and
$(S2)$ : If positive numbers $2a, b, c$ are three consecutive terms of an $A.P.$, then the lines $ax + by + c = 0$ are concurrent at $(2, -2)$.

If a line passing through the point $P(1, 1)$ makes an angle $\theta$ with the positive direction of the $x$-axis and intersects the coordinate axes at $A$ and $B$ such that $AB = OP$,where $O$ is the origin,then what is the value of $\tan\theta + \cot\theta$?

$A$ line is such that its segment between the straight lines $5x - y - 4 = 0$ and $3x + 4y - 4 = 0$ is bisected at the point $(1, 5)$,then its equation is

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 ............... $sq. \, units$.

For $a > b > c > 0$,the distance between $(1,1)$ and the point of intersection of the lines $ax + by + c = 0$ and $bx + ay + c = 0$ is less than $2\sqrt{2}$. Then: