InterviewSolution
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InterviewSolution
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Prove that the points (a, 0), (0, b) and (1, 1) are collinear if,\(\frac{1}a+\frac{1}b\) = 1 |
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Answer» Let three given points be A(a,0), B(0,b) and C(1,1). Area of the triangle having vertices (x1,y1), (x2,y2) and (x3,y3) = \(\frac{1}2\) |x1(y2 - y3)+x2(y3 - y1)+x3(y1 - y2)| Area of ∆ABC = \(\frac{1}2\) |a(b – 1) + 1(0 -b)| = \(\frac{1}2\) | ab – a –b| Here given that \(\frac{1}a+\frac{1}b\) = 1 ∴ \(\frac{a+b}{ab}\) = 1 ∴ a + b = ab Now, Area of ∆ABC = \(\frac{1}2\) | ab - (a + b)| = \(\frac{1}2\) | ab – ab| = \(\frac{1}2\) | 0 | = 0 sq. units We know that if area enclosed by three points is zero, then points are collinear. Hence, given three points are collinear. |
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