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if the pts (1,3) and (5,1) are two opposite vertices of a rectangle and the other two vertices lie on the line Y=2X+C, then the value of c is

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10 years ago 419 Points
```							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

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```
10 years ago
```							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
```
10 years ago
```							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
```
3 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 = – 4So, 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 getCoordinates 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 getCoordinates 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).
```
3 months ago
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