In an isosceles right-angled triangle,if the equation of the hypotenuse is $3x + 4y = 4$ and its opposite vertex is $(2, 2)$,then the slopes of the remaining two sides are:

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

$A$ line $L$ passes through the points $(1, 1)$ and $(2, 0)$ and another line $L'$ passes through $\left( \frac{1}{2}, 0 \right)$ and is perpendicular to $L$. Then the area of the triangle formed by the lines $L, L'$ and the $y$-axis is:

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

Three vertices of a triangle are $A(4, 3)$,$B(1, -1)$,and $C(7, k)$. The value$(s)$ of $k$ for which the centroid,orthocentre,incentre,and circumcentre of the $\Delta ABC$ lie on the same straight line is/are:

Let $m_{1}, m_{2}$ be the slopes of two adjacent sides of a square of side $a$ such that $a^{2}+11 a+3(m_{1}^{2}+m_{2}^{2})=220$. If one vertex of the square is $(10(\cos \alpha-\sin \alpha), 10(\sin \alpha+\cos \alpha))$,where $\alpha \in(0, \frac{\pi}{2})$ and the equation of one diagonal is $(\cos \alpha-\sin \alpha) x +(\sin \alpha+\cos \alpha) y =10$,then $72(\sin ^{4} \alpha+\cos ^{4} \alpha)+a^{2}-3 a+13$ is equal to.