If the three lines $x - 3y = p$,$ax + 2y = q$,and $ax + y = r$ form a right-angled triangle,then:

Let $A(h, k)$,$B(1, 1)$,and $C(2, 1)$ be the vertices of a right-angled triangle with $AC$ as its hypotenuse. If the area of the triangle is $1$ square unit,then the set of values which $k$ can take is given by

If $ad-bc \neq 0$,then the area (in sq. units) of the parallelogram formed by the lines $ax+by+2=0$,$ax+by+5=0$,$cx+dy+3=0$ and $cx+dy+7=0$ is

$A$ pair of straight lines $x^2 - 8x + 12 = 0$ and $y^2 - 14y + 45 = 0$ form a square. The coordinates of the centre of the circle inscribed in the square are

The area of the parallelogram formed by the lines $L_1 \equiv \lambda x+4 y+2=0$,$L_2 \equiv 3 x+4 y-3=0$,$L_3 \equiv 2 x+\mu y+6=0$,and $L_4 \equiv 2 x+y+3=0$,where $L_1$ is parallel to $L_2$ and $L_3$ is parallel to $L_4$,is

The equations of sides $AB$,$BC$,and $CA$ of a $\triangle ABC$ are $2x+y=0$,$x+py=q$,and $x-y=3$ respectively. If $P(2,3)$ is its orthocenter,then the value of $p+q$ equals