If the co-ordinates of the middle point of the portion of a line intercepted between coordinate axes $(3,2)$, then the equation of the line will be
$2x + 3y = 12$
$3x + 2y = 12$
$4x - 3y = 6$
$5x - 2y = 10$
Two mutually perpendicular straight lines through the origin from an isosceles triangle with the line $2x + y = 5$ . Then the area of the triangle is :
If the sum of the distances of a point from two perpendicular lines in a plane is $1$, then its locus is
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 perpendicular to $L$. Then the area of the triangle formed by the lines $L,L'$ and $y$- axis, is
A straight line through the point $(1, 1)$ meets the $x$-axis at ‘$A$’ and the $y$-axis at ‘$B$’. The locus of the mid-point of $AB$ is
Equations of diagonals of square formed by lines $x = 0,$ $y = 0,$$x = 1$ and $y = 1$are