In an isosceles triangle,the ends of its base are $(2a, 0)$ and $(0, a)$. One of its two other sides is a horizontal line (not the $X$-axis). If the third vertex is $(x_1, y_1)$,then $x_1 + y_1 =$

$A$ triangle $ABC$ lying in the first quadrant has two vertices as $A(1, 2)$ and $B(3, 1)$. If $\angle BAC = 90^{\circ}$ and $\text{ar}(\Delta ABC) = 5\sqrt{5}$ sq. units,then the abscissa of the vertex $C$ is

$A$ triangle with vertices $(4, 0), (-1, -1), (3, 5)$ is

Suppose $ABCD$ $(AB \parallel CD)$ is a trapezium such that the diagonals $AC$ and $BD$ bisect the angles $\angle DAB$ and $\angle CBA$,respectively. Then

In $\triangle ABC$,the coordinates of the vertex $A$ are $(-3, 1)$. If the equation of the median through $B$ is $2x + y - 3 = 0$ and the equation of the angle bisector of $\angle C$ is $7x - 4y - 1 = 0$,then the equation of the side $BC$ is

The points $(1,3)$ and $(5,1)$ are opposite vertices of a diagonal of a rectangle. If the other two vertices lie on the line $y=2x+c$,then one of the vertices on the other diagonal is