If $k_1 > k_2$ are the two values of $k$ such that the lines $y - 3kx + 4 = 0$ and $(2k - 1)x - (8k - 1)y - 6 = 0$ are perpendicular,then the equation of the line passing through $(k_1, k_2)$ and having the slope $\left(\frac{k_2}{k_1}\right)$ is

The equation of the line passing through the point of intersection of the lines $3x - 4y + 1 = 0$ and $5x + y - 1 = 0$ and making equal non-zero intercepts on the coordinate axes 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

$A$ line $4x + y = 1$ passes through the point $A(2, -7)$ and meets the line $BC$,whose equation is $3x - 4y + 1 = 0$,at the point $B$. The equation of the line $AC$ such that $AB = AC$ is

If $p$ and $q$ are the lengths of the perpendiculars from the origin on the lines,$x \operatorname{cosec} \alpha - y \sec \alpha = k \cot 2 \alpha$ and $x \sin \alpha + y \cos \alpha = k \sin 2 \alpha$ respectively,then $k^{2}$ 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 ............... $sq. \, units$.