Question

Prove that the points (a 0) (0 b) and (1 -1) are collinear if 1/a-1/b=1​

Answer 1

Three or more points are collinear if the slope of any two pairs of points is the same. With three points A, B and C, three pairs of points can be formed, they are AB, BC and AC. If Slope of AB = slope of BC = slope of AC, then A, B and C are collinear points...

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Answer 2

The points A(a,0), B(0,b) and (1,-1) are collinear if the area of the triangle formed by these points is zero.

$\setlength{\unitlength}{1.2cm}\begin{picture}(0,0)\linethickness{0.4mm}\qbezier(1, 0)(1,0)(3,3)\qbezier(5,0)(5,0)(3,3)\qbezier(5,0)(5,0)(1,0)\put(2.7, 3.3){\sf A(a,0)}\put(0.5, - 0.4){\sf B(0,b)}\put(4.8, -0.4){\sf C(1, - 1)}\end{picture}$

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Here, 1/a-1/b=1 (given)

b-a = ab

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★ Now, Finding area of Triangle,

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We know that,

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$\star\;{\boxed{\sf{\purple{Area\;of\; \triangle = \dfrac{1}{2} \bigg[ a(b + 1) + 0(-1 - 0) + 1(0 - b) \bigg]}}}}$

$:\implies\sf \dfrac{1}{2} \bigg[ ab + a - b \bigg]$

$:\implies\sf \dfrac{1}{2} [ ab-(b-a) ]$

$:\implies\sf \dfrac{1}{2} [ ab-ab ]$

$:\implies\sf \dfrac{1}{2} \times 0$

$:\implies{\boxed{\frak{\blue{0\;units}}}}\;\bigstar$

$\therefore\;{\underline{\sf{The\;points\;are\;collinear\;if\: \bf{\dfrac{1}{a}- \dfrac{1}{b}=1}.}}}$

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$\qquad\qquad\boxed{\underline{\underline{\bigstar \: \bf\:More\:to\:know\:\bigstar}}}$

The distance between two points A(x₁,y₁) and B (x₂,y₂) is $\sf \purple {\sqrt{(x_2 - x_1) + (y_2 - y_1)}}$

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The coordinates of point P(x,y) which divides the line segment joining A(x₁,y₁) and B (x₂,y₂) internally in the ratio m₁:m₂ are $\sf \pink {(x,y) = \bigg( \dfrac{m_2 x_1 + m_1 x_2}{m_1 + m_2} \;,\; \dfrac{m_2 y_1 + m_1 y_2}{m_1 + m_2} \bigg)}$