Let the equation of two sides of a triangle be $3x\,-\,2y\,+\,6\,=\,0$ and $4x\,+\,5y\,-\,20\,=\,0.$ If the orthocentre of this triangle is at $(1, 1),$ then the equation of its third side is

The $x -$ co-ordinates of the vertices of a square of unit area are the roots of the equation $x^2 - 3 |x| + 2 = 0$ and the $y -$ co-ordinates of the vertices are the roots of the equation $y^2 - 3y + 2 = 0$ then the possible vertices of the square is/are :

Let $\mathrm{A}(-2,-1), \mathrm{B}(1,0), \mathrm{C}(\alpha, \beta)$ and $\mathrm{D}(\gamma, \delta)$ be the vertices of a parallelogram $A B C D$. If the point $C$ lies on $2 x-y=5$ and the point $D$ lies on $3 x-2 y=6$, then the value of $|\alpha+\beta+\gamma+\delta|$ is equal to_____.

Two lines are drawn through $(3, 4)$, each of which makes angle of $45^\circ$ with the line $x - y = 2$, then area of the triangle formed by these lines is

A variable straight line passes through a fixed point $(a, b)$ intersecting the co-ordinates axes at $A\,\, \&\,\, B$. If $'O'$ is the origin then the locus of the centroid of the triangle $OAB$ is :

Let $A B C$ be an isosceles triangle in which $A$ is at $(-1,0), \angle A=\frac{2 \pi}{3}, A B=A C$ and $B$ is on the positive $\mathrm{x}$-axis. If $\mathrm{BC}=4 \sqrt{3}$ and the line $\mathrm{BC}$ intersects the line $y=x+3$ at $(\alpha, \beta)$, then $\frac{\beta^4}{\alpha^2}$ is :