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

The coordinates of the foot of the perpendicular from the point $(1, 2)$ on the line $x - 3y + 7 = 0$ are

The image of $A(1, -2)$ with respect to the straight line $L \equiv 2x - 3y + 5 = 0$ is $B$. The foot of the perpendicular from $P(-4, -1)$ onto the line joining $AB$ is:

Let $0 < \alpha < \pi/2$ be a constant angle. If $P \equiv (\cos \theta, \sin \theta)$ and $Q \equiv (\cos(\alpha - \theta), \sin(\alpha - \theta))$,how is $Q$ obtained from $P$?

If the image of the point $(3, 8)$ in the line $x + 3y = 7$ is $(\alpha, \beta)$,then $\alpha + \beta =$

The equations of the perpendicular bisectors of the sides $AB$ and $AC$ of a $\triangle ABC$ are $x-y+5=0$ and $x+2y+5=0$,respectively. If $A$ is $(1, -2)$,then the equation of the straight line $BC$ is