The ABCD is a square with sides of unit length. Points E and F are taken on sides AB and AD respectively so that AE = AF. Let P be a point inside the square ABCD. The value of ( PA )2-( PB )2+( PC )2-( PD )2 is equal to

Question

The ABCD is a square with sides of unit length. Points E and F are taken on sides AB and AD respectively so that AE = AF. Let P be a point inside the square ABCD. The value of ( PA )2-( PB )2+( PC )2-( PD )2 is equal to (A) 3 (B) 2 (C) 1 (D) 0

Tardigrade Admin · Accepted Answer

(P A)2-(P B)2+(P C)2-(P D)2 =(α2+&gamma;2) (α2+δ2)+(δ2+β2)-(&gamma;2+β2)=0

Q. The $A BC D$ is a square with sides of unit length. Points $E$ and $F$ are taken on sides $A B$ and $A D$ respectively so that $A E = A F$ . Let $P$ be a point inside the square $A BC D$ .
The value of $(P A)^{2} - (PB)^{2} + (PC)^{2} - (P D)^{2}$ is equal to

3

2

1

0

$(P A)^{2} - (PB)^{2} + (PC)^{2} - (P D)^{2}$
$= (α^{2} + γ^{2}) (α^{2} + δ^{2}) + (δ^{2} + β^{2}) - (γ^{2} + β^{2}) = 0$

Q. The ABCD is a square with sides of unit length. Points E and F are taken on sides AB and AD respectively so that AE=AF. Let P be a point inside the square ABCD. The value of (PA)2−(PB)2+(PC)2−(PD)2 is equal to

3

2

1

0

(PA)2−(PB)2+(PC)2−(PD)2 =(α2+γ2)(α2+δ2)+(δ2+β2)−(γ2+β2)=0

Q. The $A BC D$ is a square with sides of unit length. Points $E$ and $F$ are taken on sides $A B$ and $A D$ respectively so that $A E = A F$ . Let $P$ be a point inside the square $A BC D$ .
The value of $(P A)^{2} - (PB)^{2} + (PC)^{2} - (P D)^{2}$ is equal to

$(P A)^{2} - (PB)^{2} + (PC)^{2} - (P D)^{2}$
$= (α^{2} + γ^{2}) (α^{2} + δ^{2}) + (δ^{2} + β^{2}) - (γ^{2} + β^{2}) = 0$