Math, asked by milindrrampure, 1 year ago

The points A(1,3) and C(5,1) are opposite vertices of a rectangle the equation of line passing through other two vertices and of gradient 2 is

Answers

Answered by gargshreyansh2p4lh4n
70



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).
Answered by sangeetamalpedi17
2

Answer:

(2,0) and (4,4)

Step-by-step explanation:

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).

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