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 A13 Let A≡(3,3),B≡(h,0) and C≡(0,k)
Now points A,B,C are collinear, so the area of △ABC=0 ⇒12∣∣∣3h0330k3∣∣∣=0⇒|−3h+hk−3k|=0⇒3(h+k)=hk∴1h+1k=13