Prove that the points (a,0),(0,b) and (1,1) are collinear if (a1+b1=1)

Formula:Area of triangle =21[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]If the area of a triangle is zero, then the points are collinear.Given,A(a,0),B(0,b),C(1,1)Area of △ABC=21[a(b−1)+0(1−0)+1(0−b)]=21[ab−a+0−b]from given condition, we have,a1+b1=1⇒ab=a+b=(a+b)−a−b=0sq.unitsHence the given points are collinear

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Prove that the points (a,0),(0,b) and (1,1) are collinear if

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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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Prove that the points

(a, 0), (0, b)

and

(1, 1)

are collinear if

(\frac{1}{a} + \frac{1}{b} = 1)

5 mins ago

Discuss this question LIVE

5 mins ago

Text solutionVerified

Formula:

Area of triangle

= \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})]

If the area of a triangle is zero, then the points are collinear.

Given,

A (a, 0), B (0, b), C (1, 1)

Area of

△ A BC = \frac{1}{2} [a (b - 1) + 0 (1 - 0) + 1 (0 - b)]

= \frac{1}{2} [ab - a + 0 - b]

from given condition, we have,

\frac{1}{a} + \frac{1}{b} = 1

\Rightarrow ab = a + b

= (a + b) - a - b = 0

sq.units

Hence the given points are collinear

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Question Text	Prove that the points $(a, 0), (0, b)$ and $(1, 1)$ are collinear if $(\frac{1}{a} + \frac{1}{b} = 1)$
Answer Type	Text solution:1
Upvotes	150

Prove that the points (a,0),(0,b) and (1,1) are collinear if (a1​+b1​=1)

Text solutionVerified

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Prove that the points $(a, 0), (0, b)$ and $(1, 1)$ are collinear if $(\frac{1}{a} + \frac{1}{b} = 1)$