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Question

The points (3, 3), (h, 0) and (0, k) are collinear if


A

1h+1k=13

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B

1h1k=13

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C

1k1h=13

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D

1k1h=17

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Solution

The correct option is A

1h+1k=13


Let A ≡ (3, 3), B ≡ (h, 0) and C ≡ (0, k). The points A, B, C will be collinear if area of ΔABC = 0

12 [3 (0 – k) + h(k – 3) + 0(3 – 0)] = 0

12 (–3k + hk – 3h) = 0 or 3k + 3h = hk

or 3h+3k=1 or 1h+1k=13


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