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
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
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Diagonals of rectangle always intersect at midpoint.
Updated On: Apr 26, 2026
- $\left(\frac{9}{2}, 23\right)$
- $\left(9, \frac{23}{2}\right)$
- $\left(7, \frac{21}{2}\right)$
- $\left(\frac{7}{2}, 21\right)$
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The Correct Option is B
Solution and Explanation
Concept:
Diagonals of rectangle bisect each other â midpoint of opposite corners. Step 1: Find corner points. Corners: \[ (8,11), (10,12) \]
Step 2: Midpoint. \[ \left(\frac{8+10}{2}, \frac{11+12}{2}\right) = (9, \tfrac{23}{2}) \]
Step 3: Conclusion. Intersection point = $\left(9, \frac{23}{2}\right)$
Diagonals of rectangle bisect each other â midpoint of opposite corners. Step 1: Find corner points. Corners: \[ (8,11), (10,12) \]
Step 2: Midpoint. \[ \left(\frac{8+10}{2}, \frac{11+12}{2}\right) = (9, \tfrac{23}{2}) \]
Step 3: Conclusion. Intersection point = $\left(9, \frac{23}{2}\right)$
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