Question

p is any point outside the square abcd. prove that pa²-pb² = pd²-pc²​

Answer 1

Prove that pa²-pb² = pd²-pc² where P is any point outside the square ABCD

Step-by-step explanation:

Given data:

ABCD is a square with a point P located outside the square.

To prove:

PA² – PB² = PD² – PC²

Proof:

Required formula:

Distance between two points (x₁,y₁) & (x₂,y₂) = √[(x₂-x₁)² + (y₂-y₁)²] …. (i)

Step 1:

Let the side of the square be denoted as “a” units.

So, from the figure attached below, we can conclude

Coordinates of A = (0,0)

Coordinates of B = (a,0)

Coordinates of C = (a,a)

Coordinates of D = (0,a)

And, let the coordinates of point P be (x,y).

Therefore, using the formula in eq. (i) we have,

PA² = [(x-0)² + (y-0)²] = [x² + y²] …… (ii)

PB² = [(x-a)² + (y-0)²] = [(x-a)² + y²] ….. (iii)

PD² = [(x-0)² + (y-a)²] = [x² + (y-a)²] …. (iv)

PC² = [(x-a)² + (y-a)²] ….. (v)

Step 2:

Now,

Taking L.H.S. and substituting the values from eq. (ii) & (iii),

PA² – PB²

= [x² + y²] - [(x-a)² + y²]

= [x² + y² - (x-a)² - y²]

= [x² - (x-a)²] ….. (vi)

And,  

Taking R.H.S. and substituting the values from eq. (iv) & (v),  

PD² – PC²

= [x² + (y-a)²] - [(x-a)²+ (y-a)²]

= [x² + (y-a)² - (x-a)² - (y-a)²]

= [x^2 - (x-a)^2] ….. (vii)

Thus,

Comparing eq. (vi) & (vii), we get

L.H.S. = R.H.S.

PA² – PB² = PD² – PC²

Hence proved

Hope this is helpful!!!