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

Let a triangle be bounded by the lines $L _{1}: 2 x +5 y =10$; $L _{2}:-4 x +3 y =12$ and the line $L _{3}$, which passes through the point $P (2,3)$, intersect $L _{2}$ at $A$ and $L _{1}$ at $B$. If the point $P$ divides the line-segment $A B$, internally in the ratio $1: 3$, then the area of the triangle is equal to

Let $A(a, b), B(3,4)$ and $(-6,-8)$ respectively denote the centroid, circumcentre and orthocentre of a triangle. Then, the distance of the point $P(2 a+3,7 b+5)$ from the line $2 x+3 y-4=0$ measured parallel to the line $x-2 y-1=0$ is

A square of side a lies above the $x$ -axis and has one vertex at the origin. The side passing through the origin makes an angle $\alpha ,(0 < \alpha < \frac{\pi }{4})$ with the positive direction of $x$-axis. The equation of its diagonal not passing through the origin is

$ABC$ is an isosceles triangle . If the co-ordinates of the base are $(1, 3)$ and $(- 2, 7) $, then co-ordinates of vertex $A$ can be :

The equation of straight line passing through $( - a,\;0)$ and making the triangle with axes of area ‘$T$’ is