The locus of the incentre of the triangle formed by the lines $xy-4x-4y+16=0$ and $x+y=5$ is

$A$ moving line intersects the lines $x+y=0$ and $x-y=0$ at the points $A$ and $B$ respectively, such that the area of the triangle with vertices $(0,0)$, $A$, and $B$ has a constant area $C$. The locus of the mid-point of $AB$ is given by the equation:

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

For the variable $t$, the locus of the point of intersection of the lines $3tx - 2y + 6t = 0$ and $3x + 2ty - 6 = 0$ is

$ABC$ is an isosceles triangle. If the coordinates of the base are $B(1, 3)$ and $C(-2, 7)$,the coordinates of vertex $A$ can be

$p, x_1, x_2, \ldots, x_n$ and $q, y_1, y_2, \ldots, y_n$ are two arithmetic progressions with common differences $a$ and $b$ respectively. If $\alpha$ and $\beta$ are the arithmetic means of $x_1, x_2, \ldots, x_n$ and $y_1, y_2, \ldots, y_n$ respectively,then the locus of $P(\alpha, \beta)$ is