Two consecutive vertices of a rectangle of area $10 u n i t^{2}$ are $(1, 3)$ and $(- 2, - 1)$ . Other two vertices are

(\frac{- 3}{6}, \frac{21}{5}), (- \frac{18}{5}, \frac{1}{5})

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(- \frac{3}{5}, \frac{21}{5}), (- \frac{2}{5}, \frac{11}{5})

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(- \frac{2}{5}, - \frac{11}{5}), (\frac{13}{5}, \frac{9}{5})

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(\frac{13}{5}, \frac{9}{5}), (- \frac{18}{5}, \frac{1}{5})

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Solution

The correct option is A $(\frac{- 3}{6}, \frac{21}{5}), (- \frac{18}{5}, \frac{1}{5})$
$\begin{matrix} A B = \sqrt{9 + 16} = 5 u n i t A r e a o f R e c t a n g l e = 10 u n i t^{2} = b = 2 u n i t {(A D)}^{2} = {(x_{2} - 1)}^{2} + {(y_{2} - 3)}^{2} {(B C)}^{2} = {(x_{1} + 2)}^{2} + {(y_{1} + 1)}^{2} \end{matrix}$
Put value from MCQ get answer vertices satisfy.

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Two consecutive vertices of a rectangle of area 10 unit2 are (1,3) and (−2,−1). Other two vertices are

The correct option is A (−36,215),(−185,15)AB=√9+16=5unitAreaofRectangle=10unit2=b=2unit(AD)2=(x2−1)2+(y2−3)2(BC)2=(x1+2)2+(y1+1)2Put value from MCQ get answer vertices satisfy.

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