The centre of a square of side 4 units length is $(3,7)$ and one of the diagonals is parallel to the line…

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 $\left(\mathrm{x}_1, \mathrm{y}_1\right),\left(\mathrm{x}_2, \mathrm{y}_2\right),\left(\mathrm{x}_3, \mathrm{y}_3\right)$ and $\left(\mathrm{x}_4, \mathrm{y}_4\right)$ are the vertices of this square, then $\frac{y_1 y_2 y_3 y_4}{x_1 x_2 x_3 x_4}=$
  1. 81
  2. $\frac{245}{16}$
  3. 25
  4. $\frac{105}{2}$

Solution


$\begin{aligned} & =\left(3 \pm 2 \sqrt{2} \cos \frac{\pi}{4}, 7 \pm 2 \sqrt{2} \sin \frac{\pi}{4}\right)=(3 \pm 2,7 \pm 2) \\ & =(5,9) \text { and }(1,5)\end{aligned}$ (ii) Coordinates of D, B: $\begin{aligned} & \left(3 \pm 2 \sqrt{2} \cos \frac{3 \pi}{4}, 7 \pm 2 \sqrt{2} \sin \frac{3 \pi}{4}\right) \\ & =(3 \pm 2,7 \pm 2) \\ & =(1,9) \text { and }(5,5)\end{aligned}$ So, we have: $\begin{aligned} & \mathrm{A} \equiv\left(x_1, y_1\right)=(1,5) \text { and } \mathrm{B} \equiv\left(x_2, y_2\right)=(5,5) \\ & \mathrm{C} \equiv\left(x_3, y_3\right)=(5,9) \text { and } \mathrm{D} \equiv\left(x_4, y_4\right)=(1,9)\end{aligned}$ Now, $\frac{y_1 y_2 y_3 y_4}{x_1 x_2 x_3 x_4}=\frac{5 \times 5 \times 9 \times 9}{1 \times 5 \times 5 \times 1}=81$

Asked in: AP EAMCET 2023 (17 May Shift 2)

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