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प्रश्न
The point of intersection of the diagonals of the rectangle whose sides are contained in the lines x = 8, x = 10, y = 11, and y =12 is
विकल्प
\[\left(\frac{9}{2},23\right)\]
\[(9,\frac{23}{2})\]
\[\left(7,\frac{21}{2}\right)\]
\[\left(\frac{7}{2},21\right)\]
MCQ
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उत्तर
\[(9,\frac{23}{2})\]
Explanation:

Equation of diagonal AC
\[\frac{y-y_1}{y_1-y_2}=\frac{x-x_1}{x_1-x_2}\]
\[\frac{y-11}{11-12}=\frac{x-8}{8-10}\]
\[\frac{y-11}{-1}=\frac{x-8}{-2}\]
⇒ x − 2y + 14 = 0 ...(i)
The equation of the diagonal BD is
\[\frac{y-y_1}{y_1-y_2}=\frac{x-x_1}{x_1-x_2}\]
\[\frac{y-11}{11-12}=\frac{x-10}{10-8}\]
\[\frac{y-11}{-1}=\frac{x-10}{2}\]
⇒ x + 2y − 32 = 0 ...(ii)
Solving equations (i) and (ii), we get
x = 9, y = 23/2
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