The sum of the squares of the intercepts made by the line $5x - 2y = 10$ on the coordinate axes equals:

The vertex $A$ of a triangle lies on the lines $x+y=1$ and $2x+3y=6$. If the orthocentre of the triangle is $O\left(\frac{3}{7}, \frac{22}{7}\right)$,then the equation of $OA$ in the normal form is

The equations of the sides $AB$,$AC$,and $BC$ of a $\triangle ABC$ are respectively $x-3y=0$,$3x-y=0$,and $x+y+4=0$. If $P$ and $Q$ are points on the line $3x-y+k=0$ passing through $B$ such that $PB:BQ=1:1$,then $k=$

$A$ straight line passing through the point $(2, 2)$ intersects the lines $\sqrt{3}x + y = 0$ and $\sqrt{3}x - y = 0$ at points $A$ and $B$ respectively. Find the equation of the line $AB$ such that the triangle $OAB$ is an equilateral triangle,where $O$ is the origin.

In the equation $y - y_1 = m(x - x_1)$,if $m$ and $x_1$ are fixed and different lines are drawn for different values of $y_1$,then

The line $MN$ whose equation is $x-y-2=0$ cuts the $X$-axis at $M$ and the coordinates of $N$ are $(4,2)$. The line $MN$ is rotated about $M$ through $45^{\circ}$ in the anticlockwise direction. What is the equation of the line $MN$ in the new position?