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Question

If the points (3,3),(h,0) and (0,k) are collinear, then the value of 1h+1k is

A
13
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B
16
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C
19
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D
112
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Solution

The correct option is A 13
Let A(3,3),B(h,0) and C(0,k)
Now points A,B,C are collinear, so the area of ABC=0
123h0330k3=0|3h+hk3k|=03(h+k)=hk1h+1k=13

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