In what ratio does the line $2x + 3y + 7 = 0$ divide the line segment joining the points $(3, 4)$ and $(7, 8)$?

The point which divides externally the line joining the points $(a + b, a - b)$ and $(a - b, a + b)$ in the ratio $a:b$ is

The point $A$ divides the join of the points $(-5, 1)$ and $(3, 5)$ in the ratio $k : 1$. The coordinates of the points $B$ and $C$ are $(1, 5)$ and $(7, -2)$ respectively. If the area of the triangle $ABC$ is $2$ square units,then $k =$

The straight line $3x + y = 9$ divides the line segment joining the points $(1, 3)$ and $(2, 7)$ in the ratio

Find the points of trisection of the line segment joining the points $A(2, 1)$ and $B(5, 3)$.

The line $L \equiv 6x + 3y + k = 0$ divides the line segment joining the points $(3, 5)$ and $(4, 6)$ in the ratio $-5: 4$. If the point of intersection of the lines $L = 0$ and $x - y + 1 = 0$ is $P(g, h)$,then $h =$