Medium
Check whether the point $(\sqrt{2}, 4\sqrt{2})$ is a solution of the equation $x - 2y = 4$.
(D) Given equation: $x - 2y = 4$.
The point is $(\sqrt{2}, 4\sqrt{2})$,which means $x = \sqrt{2}$ and $y = 4\sqrt{2}$.
Substitute these values into the left-hand side ($L$.$H$.$S$.) of the equation:
$L$.$H$.$S$. $= x - 2y = \sqrt{2} - 2(4\sqrt{2})$
$= \sqrt{2} - 8\sqrt{2}$
$= (1 - 8)\sqrt{2} = -7\sqrt{2}$.
Since the right-hand side ($R$.$H$.$S$.) is $4$,and $-7\sqrt{2} \neq 4$,the $L$.$H$.$S$. $\neq$ $R$.$H$.$S$.
Therefore,the point $(\sqrt{2}, 4\sqrt{2})$ is not a solution of the equation $x - 2y = 4$.
The point is $(\sqrt{2}, 4\sqrt{2})$,which means $x = \sqrt{2}$ and $y = 4\sqrt{2}$.
Substitute these values into the left-hand side ($L$.$H$.$S$.) of the equation:
$L$.$H$.$S$. $= x - 2y = \sqrt{2} - 2(4\sqrt{2})$
$= \sqrt{2} - 8\sqrt{2}$
$= (1 - 8)\sqrt{2} = -7\sqrt{2}$.
Since the right-hand side ($R$.$H$.$S$.) is $4$,and $-7\sqrt{2} \neq 4$,the $L$.$H$.$S$. $\neq$ $R$.$H$.$S$.
Therefore,the point $(\sqrt{2}, 4\sqrt{2})$ is not a solution of the equation $x - 2y = 4$.
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