# The extremities of the diagonal of a square are (1,1) & (-1,-1).Obtain two vertices and equation of diagonal.

$\begin{array}{1 1}(a)\;(-1,1),(1,-1),y=-x\\(b)\;(1,1),(1,1),y=x\\(c)\;(-2,1),(1,-2),y=-2x\\(d)\;(2,1),(1,-2),y=-3x\end{array}$

Here $(1,1)$ & $(-1,-1)$ are vertices of diagonal of square which seems to be lying on the line y=x hence the line perpendicular to this y=x is y=-x.
Hence equation of the other diagonal is $y=-x$
Let $D(x_1,y_1)$ & $B(x_2,y_2)$
These points lie on the line $y=-x$ hence $y_1=-x_1,y_2=-x_2$
Now $OA=OD$ where O(0,0) midpoint of AC.
$OA=\sqrt 2$
$OD=\sqrt 2$
$\sqrt{x_1^2+y_1^2}=2$
$x_1^2+y_1^2=2$
$x_1^2+(-x_1)^2=2$
$2x_1^2=2$
$x_1=\pm 1$
So $x_1$ has two values $\pm 1$ in which $x_1=-1$ & $x_2=1$
So $y_1=1$ & $y_2=-1$
Hence $D(-1,1)$ & $B(1,-1)$
Hence (a) is the correct answer.
answered Feb 5, 2014