The equation of one of the sides of an isosceles right-angled triangle,whose hypotenuse is $3x + 4y = 4$ and the vertex opposite to the hypotenuse is $(2, 2)$,is:

The coordinates of the vertices $A$ and $B$ of an isosceles triangle $ABC$ $(AC = BC)$ are $(-2, 3)$ and $(2, 0)$ respectively. $A$ line parallel to $AB$ and having a $y$-intercept equal to $\frac{43}{12}$ passes through $C$. Then the coordinates of $C$ are:

$L_1 \equiv ax-3y+5=0$ and $L_2 \equiv 4x-6y+8=0$ are two parallel lines. If $p, q$ are the intercepts made by $L_1=0$ and $m, n$ are the intercepts made by $L_2=0$ on the $X$ and $Y$ coordinate axes respectively,then the equation of the line passing through the points $(p, q)$ and $(m, n)$ is

If $l, m, n$ are in arithmetic progression,then the straight line $lx + my + n = 0$ will always pass through the point:

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

The point on the line $4x - y - 2 = 0$ which is equidistant from the points $(-5, 6)$ and $(3, 2)$ is: