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If the points (h, o), (a, b) and (o, k) lie on a line, then?
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a + b = 1 h k -
a + b = 1 k h -
h + k = 1 a b -
a - b = 1 h k
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Correct Option: A
We know that when three points are collinear than area of triangle is zero.
| ar∆ = |  | h | 0 | 1 | | |||
| a | b | 1 | ||||||
| 2 | a | b | 1 |
| ⇒ | [h(b - k) + 1)ak)] = 0 | |
| 2 |
⇒ bh – hk + ak = 0
ak + bh = hk
Dividing both sides by hk, we get
| + | = 1 | |||
| hk | hk |
| + | = 1 | |||
| h | k |
