The co-ordinates of the orthocentre of the triangle bounded by the lines, $4x - 7y + 10 = 0; x + y=5$ and $7x + 4y = 15$ is :

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

If the extremities of the base of an isosceles triangle are the points $(2a,0)$ and $(0,a)$ and the equation of one of the sides is $x = 2a$, then the area of the triangle is

$P (x, y)$ moves such that the area of the triangle formed by $P, Q (a , 2 a)$ and $R (- a, - 2 a)$ is equal to the area of the triangle formed by $P, S (a, 2 a)\,\,\, \&\,\, \,T (2 a, 3 a)$. The locus of $'P'$ is a straight line given by :

The sides $AB,BC,CD$ and $DA$ of a quadrilateral are $x + 2y = 3,\,x = 1,$ $x - 3y = 4,\,$ $\,5x + y + 12 = 0$ respectively. The angle between diagonals $AC$ and $BD$ is ......$^o$

Show that the area of the triangle formed by the lines

$y=m_{1} x+c_{1}, y=m_{2} x+c_{2}$ and $x=0$ is $\frac{\left(c_{1}-c_{2}\right)^{2}}{2\left|m_{1}-m_{2}\right|}$.