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
$6x + y - 32 = 0$
$6x - y + 32 = 0$
$x + 6y - 32 = 0$
$6x - y - 32 = 0$
Locus of the image of point $ (2,3)$ in the line $\left( {2x - 3y + 4} \right) + k\left( {x - 2y + 3} \right) = 0,k \in R$ is a:
If two vertices of a triangle are $(5, -1)$ and $( - 2, 3)$ and its orthocentre is at $(0, 0)$, then the third vertex
Without using the Pythagoras theorem, show that the points $(4,4),(3,5)$ and $(-1,-1)$ are vertices of a right angled triangle.
$A(-1, 1)$, $B(5, 3)$ are opposite vertices of a square in $xy$-plane. The equation of the other diagonal (not passing through $(A, B)$ of the square is given by