The centre of a square of side 4 units length | vedclass

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(A) The centre of the square is $P(3,7)$ and the side length is $4$. The length of the diagonal is $4\sqrt{2}$,so the distance from the centre to each

Vedclass · Accepted Answer

$81$

Vedclass · Answer

(A) The centre of the square is $P(3,7)$ and the side length is $4$. The length of the diagonal is $4\sqrt{2}$,so the distance from the centre to each vertex is $r = \frac{4\sqrt{2}}{2} = 2\sqrt{2}$.One diagonal is parallel to $y=x$,so its slope is $m_1 = 1$. The angle it makes with the $x$-axis is $	heta_1 = 45^\circ$ or $\frac{\pi}{4}$.The other diagonal is perpendicular to the first,so its slope is $m_2 = -1$,and its angle is $	heta_2 = 135^\circ$ or $\frac{3\pi}{4}$.Using the parametric form of a line,the coordinates of the vertices are $(x, y) = (3 + r \cos 	heta, 7 + r \sin 	heta)$.For the first diagonal $(	heta = 45^\circ)$: $(3 \pm 2\sqrt{2} \cos 45^\circ, 7 \pm 2\sqrt{2} \sin 45^\circ) = (3 \pm 2, 7 \pm 2)$,giving points $(5,9)$ and $(1,5)$.For the second diagonal $(	heta = 135^\circ)$: $(3 \pm 2\sqrt{2} \cos 135^\circ, 7 \pm 2\sqrt{2} \sin 135^\circ) = (3 \mp 2, 7 \pm 2)$,giving points $(1,9)$ and $(5,5)$.The vertices are $(1,5), (5,5), (5,9), (1,9)$.Thus,$\frac{y_1 y_2 y_3 y_4}{x_1 x_2 x_3 x_4} = \frac{5 	imes 5 	imes 9 	imes 9}{1 	imes 5 	imes 5 	imes 1} = \frac{2025}{25} = 81$.

The centre of a square of side $4$ units length is $(3,7)$ and one of the diagonals is parallel to the line $y=x$. If $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ and $(x_4, y_4)$ are the vertices of this square,then $\frac{y_1 y_2 y_3 y_4}{x_1 x_2 x_3 x_4}=$

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