Question

The points \( (1, 3) \) and \( (5, 1) \) are two opposite vertices of a rectangle. The other two vertices lie on the line \[ y = 2x + c, \quad \text{where} \quad c \text{ is the constant}, \quad \text{then the co-ordinates of the other two vertices are} \]

• A. \( (4, 4), (2, 0) \)
• B. \( (4, 4), (1, 0) \)
• C. \( (2, 0), (4, 1) \)
• D. \( (2, 0), (1, -1) \)

Hint:

In geometry problems involving rectangles, use the property that the diagonals bisect each other at the mid-point to find the coordinates of unknown vertices.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The diagonals of a rectangle bisect each other. Thus, the midpoint of the given vertices is also the midpoint of the other two vertices. These other vertices lie on a specific line.
Step 2: Key Formula or Approach:
1. Find the midpoint \( M \) of \( (1, 3) \) and \( (5, 1) \).
2. Since \( M \) lies on the line \( y = 2x + c \), find \( c \).
3. The distance from \( M \) to any vertex is equal.
Step 3: Detailed Explanation:
Midpoint \( M = \left( \frac{1+5}{2}, \frac{3+1}{2} \right) = (3, 2) \).
The line \( y = 2x + c \) passes through \( (3, 2) \):
\( 2 = 2(3) + c \Rightarrow c = -4 \).
The line is \( y = 2x - 4 \).
Let the other vertices be \( (x, 2x-4) \).
Distance between given opposite vertices (diagonal length \( D \)):
\( D = \sqrt{(5-1)^2 + (1-3)^2} = \sqrt{16+4} = \sqrt{20} = 2\sqrt{5} \).
Distance from center \( M(3,2) \) to other vertices is \( D/2 = \sqrt{5} \).
\( (x-3)^2 + (2x-4-2)^2 = (\sqrt{5})^2 \)
\( (x-3)^2 + (2x-6)^2 = 5 \)
\( (x-3)^2 + 4(x-3)^2 = 5 \Rightarrow 5(x-3)^2 = 5 \Rightarrow (x-3)^2 = 1 \).
Case 1: \( x-3 = 1 \Rightarrow x = 4 \). Then \( y = 2(4)-4 = 4 \). Point is \( (4,4) \).
Case 2: \( x-3 = -1 \Rightarrow x = 2 \). Then \( y = 2(2)-4 = 0 \). Point is \( (2,0) \).
Step 4: Final Answer:
The other two vertices are \( (4, 4) \) and \( (2, 0) \).