$Δ P Q R$ is right angled at Q. S is the mid-point of QR and $P R^{2} = a (P S)^{2} + b (P Q)^{2}$

Then a + b is equal to:

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Solution

The correct option is B
1

In $Δ P Q R, P R^{2} = P Q^{2} + Q R^{2}$ ........ (1)

QR=2 QS, (Since S is the mid-point QR)

Then (1) becomes

$P R^{2} = P Q^{2} + (2 Q S)^{2}$

$P R^{2} = P Q^{2} + 4 (Q S)^{2}$ ........ (2)

From $Δ P Q S, Q S^{2} = P S^{2} - P Q^{2}$

Putting this in (2) gives,

$P R^{2} = P Q^{2} + 4 (P S^{2} - P Q^{2})$

$P R^{2} = P Q^{2} + 4 (P S)^{2} - 4 (P Q)^{2}$

$P R^{2} = 4 (P S)^{2} - 3 (P Q)^{2}$

So, a = 4 and b = -3

a + b = 4 - 3 = 1

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