Dear Deva
diagonals of rectangle bisect each other.let ABCD is rectangle. and A=(1,3), C=(5,1)
Mid point of diagonal AC is (3,2)
mid point of diagonal BD will also be (3,2) and given line is equation of BD
(3,2) will satisfy the equation
put y=2 and x=3
2=6+c
c=-4
All the best.
AKASH GOYAL
AskiitiansExpert-IITD
 
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LET ABCD is the rectangle  let A,C are (1,3) , (5,1) then         mid point of line AC is (3,2)         now mid point of line BD is same so line passing through B,D will pass mid point & (3,2) will satisfy the equation of line..  Y = 2X+C at (3,2) 2 = 3*2 + C C= -4
						

							Diagonals bisectors each other so mid point of given diagonal would also be mid point of other i.e, lies on equation and would satisfy so we get close =, - 4 on putting the point on line
						

							
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).