The diagonals of a parallelogram $ABCD$ are along the lines $x+3y=4$ and $6x-2y=7$. Then $ABCD$ must be a

In a triangle $ABC,$ side $AB$ has the equation $2x + 3y = 29$ and the side $AC$ has the equation $x + 2y = 16.$ If the mid-point of $BC$ is $(5, 6),$ then the equation of $BC$ is:

The equations of the sides $AB$,$BC$,and $CA$ of a triangle $ABC$ are $2x + y = 0$,$x + py = 21a$ $(a \neq 0)$,and $x - y = 3$ respectively. Let $P(2, a)$ be the centroid of $\triangle ABC$. Then $(BC)^2$ is equal to $........$

Suppose $ABCD$ is a trapezium whose sides and height are integers and $AB$ is parallel to $CD$. If the area of $ABCD$ is $12$ and the sides are distinct,then $|AB-CD|$ is:

The equation of the base $BC$ of an equilateral triangle is $3x + 4y = 1$ and the vertex $A$ is $(-3, 2)$. Then the area of the triangle is-

What is the angle between the diagonals of the parallelogram formed by the lines $ℓx + my + n = 0$,$ℓx + my + n' = 0$,$mx + ℓy + n = 0$,and $mx + ℓy + n' = 0$?