The locus of a point so that sum of its distance from two given perpendicular lines is equal to $2$ unit in first quadrant, is
$x + y + 2 = 0$
$x + y = 2$
$x - y = 2$
None of these
In a triangle $ABC,$ side $AB$ has the equation $2 x + 3 y = 29$ and the side $AC$ has the equation , $x + 2 y = 16$ . If the mid - point of $BC$ is $(5, 6)$ then the equation of $BC$ 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
A pair of straight lines $x^2 - 8x + 12 = 0$ and $y^2 - 14y + 45 = 0$ are forming a square. Co-ordinates of the centre of the circle inscribed in the square are
The point moves such that the area of the triangle formed by it with the points $(1, 5)$ and $(3, -7)$ is $21$ sq. unit. The locus of the point is
The diagonals of a parallelogram $PQRS$ are along the lines $x + 3y = 4$ and $6x - 2y = 7$. Then $PQRS$ must be a