The equation of the straight line which passes through the point $(1, -2)$ and cuts off equal intercepts from the axes is:

$A$ line $L$ intersects the lines $3x - 2y - 1 = 0$ and $x + 2y + 1 = 0$ at the points $A$ and $B$. If the point $(1, 2)$ bisects the line segment $AB$ and $\frac{x}{a} + \frac{y}{b} = 1$ is the equation of the line $L$,then $a + 2b + 1 = $

Two points $(a, 0)$ and $(0, b)$ are joined by a straight line. Another point on this line is:

Prove that the equation of the line passing through the point $(x_{1}, y_{1})$ and parallel to the line $Ax + By + C = 0$ is $A(x - x_{1}) + B(y - y_{1}) = 0$.

The equation of a given straight line is $\frac{x-x_1}{\cos \theta}=\frac{y-y_1}{\sin \theta}=\gamma$. If the equation of the line perpendicular to the given line and passing through $(\alpha, \beta)$ is $\frac{x}{a}+\frac{y}{b}=1$,then $\frac{b}{a}$ is equal to

Find the equation of the line passing through the point $(1, -2)$ and making equal intercepts on the axes.