Question

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

Options

• \[\left(\frac{9}{2},23\right)\]

• \[(9,\frac{23}{2})\]

• \[\left(7,\frac{21}{2}\right)\]

• \[\left(\frac{7}{2},21\right)\]

Answer 1

\[(9,\frac{23}{2})\]

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