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in Mathematics by (64.7k points)

Consider the points A(–5, –1), B(–1, 0), C(1, 2) and D(1, 3). Let P be a point such that d = PA2 + PB2 + PC2 + PD2 . The least possible value of d is : 

(1) 28 

(2) 30 

(3) 32 

(4) 34 

1 Answer

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by (58.3k points)
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Best answer

Correct option (4) 34

Explanation:

PA2 + PB2 + PC2 + PD2 = 16 + 4 + 1 + 4 + 1 + 4 + 4 = 34 

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