The equation of a line passing through $(1, -2)$ and perpendicular to the line $3x - 5y + 7 = 0$ is

The foot of the perpendicular from the point $(2,4)$ upon the line $x+y=4$ is:

Let $a$ and $b$ be any two numbers satisfying $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{4}$. Then,the foot of the perpendicular from the origin on the variable line $\frac{x}{a} + \frac{y}{b} = 1$ lies on

If $(\alpha, \beta)$ is the image of the point $(3, -4)$ with respect to the line $4x - y - 1 = 0$,then the value of $\beta - \alpha$ is equal to

The distance between the point $(2, 1)$ and the image of the point $(3, -1)$ with respect to the line $2x + y - 1 = 0$ is

If $(-2, 6)$ is the image of the point $(4, 2)$ with respect to the line $L = 0$,then $L$ is equal to