The equations of two sides $\mathrm{AB}$ and $\mathrm{AC}$ of a triangle $\mathrm{ABC}$ are $4 \mathrm{x}+\mathrm{y}=14$ and $3 \mathrm{x}-2 \mathrm{y}=5$, respectively. The point $\left(2,-\frac{4}{3}\right)$ divides the third side $\mathrm{BC}$ internally in the ratio $2: 1$. The equation of the side $\mathrm{BC}$ is :

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 :

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 :

The line $\frac{x}{a} + \frac{y}{b}=1$ moves in such a way that $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{2c^2},$ where $a, b, c \in R_0$ and $c$ is constant, then locus of the foot of the perpendicular from the origin on the given line is –

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_____.