What is the distance of the point $(6, 8)$ from the $x$-axis?

The point on the $y$-axis equidistant from the points $(3, 2)$ and $(-1, 3)$ is

Three vertices of a parallelogram taken in order are $(-1, -6)$,$(2, -5)$ and $(7, 2)$. The fourth vertex is

The distance of the midpoint of the line segment joining the points $(a \sin \theta, 0)$ and $(0, a \cos \theta)$ from the origin is:

The points which trisect the line segment joining the points $(0, 0)$ and $(9, 12)$ are

If the point $M(x, y)$ divides the line segment joining $A(b \cos \alpha, b \sin \alpha)$ and $B(a \cos \beta, a \sin \beta)$ in the ratio $b:a$,then $x \cos \frac{\alpha + \beta}{2} + y \sin \frac{\alpha + \beta}{2} = $