Question

In a rectangle ABCD, P is any interior point. then prove that PA^2+PC^2=PB^2+PD^2

Answer 1

[Hint: Take A as the origin and AB and AD as x and y - axis respectively. Let AB = a, AD = b]

Let us assume A as the origin (0, 0) and AB and AD as x and y axis with length a and b units.

Then we get points B to be (a, 0), D to be (0, b) and C to be (a, b).

Let us assume P(x, y) be any point in a plane of the rectangle.

We need to prove PA2 + PC2 = PB2 + PD2.

We know that distance between two points (x1, y1) and (x2, y2) is

This step is on image above.

Let us assume L.H.S,

⇒ PA2 + PC2 = ((x - 0)2 + (y - 0)2) + ((x - a)2 + (y - b)2)

⇒ PA2 + PC2 = x2 + y2 + x2 - 2ax + a2 + y2 - 2by + b2

⇒ PA2 + PC2 = (x2 - 2ax + a2 + y2) + (x2 + y2 - 2by + b2)

⇒ PA2 + PC2 = ((x - a)2 + (y - 0)2) + ((x - 0)2 + (y - b)2)

⇒ PA2 + PC2 = PB2 + PD2

⇒ L.H.S = R.H.S

∴ Thus proved.

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Answer 2

Step-by-step explanation:

Cytoplasm is a thick solution that fills each cell and is enclosed by the cell membrane. It is mainly composed of water, salts, and proteins. ... All of the organelles in eukaryotic cells, such as the nucleus, endoplasmic reticulum, and mitochondria, are located in the cytoplasm.