The system of equations $4x + 6y = 5$ and $8x + 12y = 10$ has

The points $(1, 1)$,$(0, \sec^2 \theta)$,and $(\csc^2 \theta, 0)$ are collinear for:

$A$ straight line $L$ is perpendicular to the line $5x - y = 1$ and the area of the triangle formed by the line $L$ and the coordinate axes is $5$ square units. The equation of the line $L$ can be

One side of a rectangle lies along the line $4x + 7y + 5 = 0.$ Two of its vertices are $(-3, 1)$ and $(1, 1).$ Then the equations of the other three sides are

$A$ straight line $L$ passes through the point of intersection of the lines $x - y + 1 = 0$ and $2x + y - 7 = 0$. If $L$ intersects the positive $x$-axis at $A(a, 0)$ and the positive $y$-axis at $B(0, b)$, then the minimum area of the triangle $OAB$ (where $O$ is the origin) is ....

$A$ straight line $x-2y-4=0$ is shifted parallel to it by $3$ units away from the origin and then rotated by an angle of $30^{\circ}$ in the anti-clockwise direction. If the slope of the new line formed is $m$,then the integral part of $m$ is