For what value of $y$ does the triangle with vertices $A(2, 7)$,$B(4, y)$,and $C(-2, 6)$ form a right angle at $A$?

Find the incenter of the triangle formed by the axes and the line $\frac{x}{a} + \frac{y}{b} = 1$.

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

The centre of a square of side $4$ units length is $(3,7)$ and one of the diagonals is parallel to the line $y=x$. If $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ and $(x_4, y_4)$ are the vertices of this square,then $\frac{y_1 y_2 y_3 y_4}{x_1 x_2 x_3 x_4}=$

If the points $(1, 1)$,$(-1, -1)$,and $(-\sqrt{3}, k)$ are the vertices of an equilateral triangle,then what is the value of $k$?

Let $ABC$ be a triangle with $A(-3, 1)$ and $\angle ACB = \theta$,where $0 < \theta < \frac{\pi}{2}$. If the equation of the median through $B$ is $2x + y - 3 = 0$ and the equation of the angle bisector of $C$ is $7x - 4y - 1 = 0$,then $\tan \theta$ is equal to: