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 $A$?

The equation of the base of an equilateral triangle is $x+y=2$ and its opposite vertex is $(2,1)$. If $m_1, m_2$ are the slopes of the other two sides and the length of its side is $a$,then $|m_1-m_2|+a \sqrt{2}=$

In an isosceles triangle $ABC$,the vertex $A$ is $(6,1)$ and the equation of the base $BC$ is $2x + y = 4$. Let the point $B$ lie on the line $x + 3y = 7$. If $(\alpha, \beta)$ is the centroid of $\triangle ABC$,then $15(\alpha + \beta)$ is equal to

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:

$A$ quadrilateral has distinct integer side lengths. If the second-largest side has length $10$,then the maximum possible length of the largest side is

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