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Match the items of column-1 with items of column-2 and choose the correct option.
Column - I Column - II
(A) If the distance between the $(2, - 2)$ and $(- 1, k)$ is 5, then k = (1) –6
(B) The perimeter of a square with vertices $(1, 7), (4, 2), (- 1, - 1), (- 4, 4)$ is (2) –5
(C) The distance between $(x, 7) a n d (1, 15)$ is 10 then x = (3) –9
(D) The distance between $(2, - 3) a n d (10, y)$ is 10 then y = (4) $4 \sqrt{34}$

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A→4, B→1, C→2, D→3

A→1, B→4, C→2, D→3

A→3, B→2, C→4, D→1

A→3, B→1, C→2, D→4

answer is B.

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Detailed Solution

$\sqrt{{(- 1 - 2)}^{2} + {(k + 2)}^{2}} = 5$

$\Rightarrow 9 + {(k + 2)}^{2} = 25$

$\Rightarrow {(k + 2)}^{2} = 16$

$\Rightarrow k + 2 = \pm 4$

$\Rightarrow k = 2 o r - 6$

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$A B = \sqrt{3^{2} + 5^{2}}$

$= \sqrt{9 + 25}$

$= \sqrt{34}$

$∴$ Perimeter of a square $= 4 s = 4 \sqrt{34}$

$\sqrt{{(1 - x)}^{2} + {(15 - 7)}^{2}} = 10$

$\Rightarrow {(1 - x)}^{2} + 64 = 100$

$\Rightarrow {(1 - x)}^{2} = 36$

$\Rightarrow (1 - x) = \pm 6$

$\Rightarrow 1 - x = 6 (o r) 1 - x = - 6$

$\Rightarrow x = - 5 (o r) x = 7$

$\sqrt{8^{2} + {(y + 3)}^{2}} = 10$

$\Rightarrow 64 + {(y + 3)}^{2} = 100$

$\Rightarrow {(y + 3)}^{2} = 36$

$\Rightarrow y + 3 = \pm 6$

$\Rightarrow y = 3 (o r) y = - 9$

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