Consider the points A(–5, –1), B(–1, 0), C(1, 2) and D(1, 3). Let P be a point such that d = PA^2 + PB^2 + PC^2 + PD^2 .

Question 1

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

Question 2

Correct option (4) 34

Explanation:

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

kratos · Answer 1 · 2020-04-18T12:57:59+0000

Correct option (4) 34

Explanation:

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

Consider the points A(–5, –1), B(–1, 0), C(1, 2) and D(1, 3). Let P be a point such that d = PA^2 + PB^2 + PC^2 + PD^2 .

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Consider the points A(–5, –1), B(–1, 0), C(1, 2) and D(1, 3). Let P be a point such that d = PA^2 + PB^2 + PC^2 + PD^2 .

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