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Q.
One of the diagonals of a square is the portion of the line $x / 2+y / 3=2$ intercepted between the axes. Then the extremities of the other diagonal are
Straight Lines
Solution:
The extremities of the given diagonal are $(4,0)$ and $(0,6)$. Hence, the slope of this diagonal is $-3 / 2$ and the slope of other diagonal is $2 / 3$. The equation of the other diagonal is
$\frac{x-2}{3 / \sqrt{13}}=\frac{y-3}{2 / \sqrt{13}}=r$
For the extremities of the diagonal, $r=\pm \sqrt{13}$. Hence,
$x-2=\pm 3, y-3=\pm 2 $
$x=5,-1 \text { and } y=5,1$
Therefore, the extremities of the diagonal are $(5,5)$ and $(-1,1)$.