The medians $AD$ and $BE$ of a triangle with vertices $A\;(0,\;b),\;B\;(0,\;0)$ and $C\;(a,\;0)$ are perpendicular to each other, if

The origin and the points where the line $L_1$ intersect the $x$ -axis and $y$ -axis are vertices of right angled triangle $T$ whose area is $8$. Also the line $L_1$ is perpendicular to line $L_2$ : $4x -y = 3$, then perimeter of triangle $T$ is -

The triangle formed by ${x^2} - 9{y^2} = 0$ and $x = 4$ is

One diagonal of a square is along the line $8x - 15y = 0$ and one of its vertex is $(1, 2)$ Then the equation of the sides of the square passing through this vertex, are

If two vertices of a triangle are $(5, -1)$ and $( - 2, 3)$ and its orthocentre is at $(0, 0)$, then the third vertex

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 :