The point whose abscissa is equal to its ordinate and which is equidistant from the points $(1, 0)$ and $(0, 3)$ is

If $a, b, c$ are in harmonical progression,then the line $bcx + cay + ab = 0$ passes through a fixed point whose coordinates are:

If the point $(x, y)$ is equidistant from the points $(a + b, b - a)$ and $(a - b, a + b)$,then:

The perpendicular bisector of the line segment joining the points $P(1, 4)$ and $Q(k, 3)$ has a $y$-intercept of $-4$. Then a possible value of $k$ among the following is:

The point moves such that the area of the triangle formed by it with the points $(1, 5)$ and $(3, -7)$ is $21$ sq. units. The locus of the point is:

Let the algebraic sum of the perpendicular distances from the points $A(2, 0)$,$B(0, 2)$,and $C(1, 1)$ to a variable line be zero. Then,all such lines: