If the line $3x + 3y -24 = 0$ intersects the $x-$ axis at the point $A$ and the $y-$ axis at the point $B$, then the incentre of the triangle $OAB$, where $O$ is the origin, is

If the straight lines $x + 3y = 4,\,\,3x + y = 4$ and $x +y = 0$ form a triangle, then the triangle is

Consider the lines $L_1$ and $L_2$ defined by

$L _1: x \sqrt{2}+ y -1=0$ and $L _2: x \sqrt{2}- y +1=0$

For a fixed constant $\lambda$, let $C$ be the locus of a point $P$ such that the product of the distance of $P$ from $L_1$ and the distance of $P$ from $L_2$ is $\lambda^2$. The line $y=2 x+1$ meets $C$ at two points $R$ and $S$, where the distance between $R$ and $S$ is $\sqrt{270}$.

Let the perpendicular bisector of $RS$ meet $C$ at two distinct points $R ^{\prime}$ and $S ^{\prime}$. Let $D$ be the square of the distance between $R ^{\prime}$ and $S ^{\prime}$.

($1$) The value of $\lambda^2$ is

($2$) The value of $D$ is

Show that the area of the triangle formed by the lines

$y=m_{1} x+c_{1}, y=m_{2} x+c_{2}$ and $x=0$ is $\frac{\left(c_{1}-c_{2}\right)^{2}}{2\left|m_{1}-m_{2}\right|}$.

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

The ends of the base of an isosceles triangle are at $(2a,\;0)$ and $(0,\;a).$ The equation of one side is $x=2a$ The equation of the other side is