The area of the triangle formed by the line $x \sin \alpha + y \cos \alpha = \sin 2\alpha$ and the coordinate axes is

If a diagonal of a square is along the line $8x - 15y = 0$ and one of its vertices is $(1, 2)$,then the equations of the sides of the square passing through this vertex are

$A$ square is formed by the lines $x=0, y=0, x=1, y=1$. Then,the equations of its diagonals will be

Let $\left(5, \frac{a}{4}\right)$ be the circumcenter of a triangle with vertices $A(a, -2)$,$B(a, 6)$,and $C\left(\frac{a}{4}, -2\right)$. Let $\alpha$ denote the circumradius,$\beta$ denote the area,and $\gamma$ denote the perimeter of the triangle. Then $\alpha + \beta + \gamma$ is

$A$ straight line $4x+y-1=0$ passing through the point $A(2,-7)$ meets the line $BC$ whose equation is $3x-4y+1=0$ at the point $B$. Then the equation of the line $AC$ such that $AB=AC$,is

Points inside the triangle with vertices at $(1,3), (5,0)$ and $(-1,2)$ must necessarily satisfy which of the following conditions?