The points $A(2a, 4a)$,$B(2a, 6a)$,and $C(2a + \sqrt{3}a, 5a)$,where $a > 0$,are the vertices of:

The diagonals $AC$ and $BD$ of a rhombus $ABCD$ intersect at the point $(3,4)$. If $BD=2 \sqrt{2}$,$A=(1,2)$,$B=(\alpha, \beta)$,$D=(\gamma, \delta)$ and $\alpha < \delta < \gamma < \beta$,then $\beta+\gamma-\delta=$

When are the points with coordinates $(2a, 3a)$,$(3b, 2b)$,and $(c, c)$ collinear?

If the line $3x + 4y - 24 = 0$ intersects $X$ and $Y$ axes at points $A$ and $B$ respectively,then the incenter of the triangle $OAB$,where $O$ is the origin,is:

The orthocentre of the triangle formed by the lines $4x - 7y + 10 = 0$,$x + y = 5$,and $7x + 4y = 15$ is

$L \equiv 7x - y + 8 = 0$ is one of the diagonals of a square for which $(-4, 5)$ and $(3, 4)$ are two vertices. Find the coordinates of the two vertices lying on the diagonal $L = 0$.