If the three lines $x - 3y = p, ax + 2y = q$ and $ax + y = r$ form a right-angled triangle then
$a^2 -9a + 18 =0$
$a^2 -6a-12=0$
$a^2 -6a- 18=0$
$a^2 -9a+ 12 =0$
The vertex of a right angle of a right angled triangle lies on the straight line $2x + y - 10 = 0$ and the two other vertices, at points $(2, -3)$ and $(4, 1)$ then the area of triangle in sq. units is
Locus of the points which are at equal distance from $3x + 4y - 11 = 0$ and $12x + 5y + 2 = 0$ and which is near the origin is
The area of a parallelogram formed by the lines $ax \pm by \pm c = 0$, is
The four points whose co-ordinates are $(2, 1), (1, 4), (4, 5), (5, 2)$ form :
If a variable line drawn through the point of intersection of straight lines $\frac{x}{\alpha } + \frac{y}{\beta } = 1$and $\frac{x}{\beta } + \frac{y}{\alpha } = 1$ meets the coordinate axes in $A$ and $B$, then the locus of the mid point of $AB$ is