Two lines are drawn through $(3, 4)$, each of which makes angle of $45^\circ$ with the line $x - y = 2$, then area of the triangle formed by these lines is

The $x -$ co-ordinates of the vertices of a square of unit area are the roots of the equation $x^2 - 3 |x| + 2 = 0$ and the $y -$ co-ordinates of the vertices are the roots of the equation $y^2 - 3y + 2 = 0$ then the possible vertices of the square is/are :

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 $\frac{\left(c_{1}-c_{2}\right)^{2}}{2\left|m_{1}-m_{2}\right|}$.

Let $A (-3, 2)$ and $B (-2, 1)$ be the vertices of a triangle $ABC$. If the centroid of this triangle lies on the line $3x + 4y + 2 = 0$, then the vertex $C$ lies on the line

If the straight line $ax + by + c = 0$ always passes through $(1, -2),$ then $a, b, c$ are