Question:medium

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

Show Hint

Diagonals of rectangle always intersect at midpoint.
Updated On: May 14, 2026
  • $\left(\frac{9}{2}, 23\right)$
  • $\left(9, \frac{23}{2}\right)$
  • $\left(7, \frac{21}{2}\right)$
  • $\left(\frac{7}{2}, 21\right)$
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
The given lines form a rectangle whose sides are parallel to the coordinate axes. A key property of any rectangle is that the intersection point of its diagonals is exactly its geometric center. For a rectangle aligned with the axes, this center is simply the midpoint of its horizontal and vertical boundaries. Step 2: Key Formula or Approach:
The midpoint formula for a line segment between $(x_1, y_1)$ and $(x_2, y_2)$ is $M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$. For an axis-aligned rectangle bounded by $x=x_1, x=x_2$ and $y=y_1, y=y_2$, the center is exactly $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$. Step 3: Detailed Explanation:
The vertical boundaries of the rectangle are given by $x = 8$ and $x = 10$. The horizontal boundaries of the rectangle are given by $y = 11$ and $y = 12$. The x-coordinate of the center is the average of the x-boundaries: \[ x_{\text{center}} = \frac{8 + 10}{2} = \frac{18}{2} = 9 \] The y-coordinate of the center is the average of the y-boundaries: \[ y_{\text{center}} = \frac{11 + 12}{2} = \frac{23}{2} \] Therefore, the point of intersection of the diagonals is the center point $(9, \frac{23}{2})$. Step 4: Final Answer:
The point of intersection is $\left(9, \frac{23}{2}\right)$.
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