ankit singh
Last Activity: 4 Years ago
ABCD is a rectangular.
Let A(1, 3), B(x1, y1), C(5, 1) and D(x2, y2) be the vertices of the rectangular.
We know that, diagonals of rectangular bisect each other.
Let O be the point of intersection of diagonal AC and BD.
∴ Mid point of AC = Mid point BD.
Now, O(3, 2) lies on y = 2x + c.
∴ 2 = 2 × 3 + c
⇒ c = 2 – 6 = – 4
So, the value of c is – 4.
(x1, y1) lies on y = 2x – 4.
∴ y1 = 2x1 – 4 ...(2)
(x2, y2) lies on y = 2x – 4
∴ y2 = 2x2 – 4 ...(3)
Coordinates of B = (x1, 2x1 – 4)
Coordinates of D = (x2, 2x2 – 4)
AD ⊥ AB,
∴ Slope of AD × Slope of AB = – 1.
When x1 = 4 and x2 = 2, we get
Coordinates of B = (x1, 2x1 – 4) = (4, 2 × 4 – 4) = (4, 4)
Coordinates of D = (x2, 2x2 – 4) = (2, 2 × 2 – 4) = (2, 0)
When x1 = 2 and x2 = 4, we get
Coordinates of B = (x1, 2x1– 4) = (4, 2 × 4 – 4) = (2, 0)
Coordinates of D = (x2, 2x2 – 4) = (4, 2 × 4 – 4) = (4, 4)
Thus, the other two vertices of the rectangle are (2, 0) and (4, 4).