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Collinear Points

The points wi...

Question

The points with the coordinates $(2 a, 3 a), (3 b, 2 b)$ and $(c, c)$ are collinear :

for no value of

a, b, c

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for all value of

a, b, c

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a, \frac{c}{5}

are in

H P

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a, \frac{2 c}{5}, b

are in

H P

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Solution

The correct option is D if $a, \frac{2 c}{5}, b$ are in $H P$
Given,

$(2 a, 3 a), (3 b, 2 b), (c, c)$

We are using slope formula,

$\frac{2 b - 3 a}{3 b - 2 a} = \frac{c - 2 b}{c - 3 b} = \frac{c - 3 a}{c - 2 a}$

considering any of the 2 equations, we get,

$\frac{c - 2 b}{c - 3 b} = \frac{c - 3 a}{c - 2 a}$

$(c - 2 a) (c - 2 b) = (c - 3 b) (c - 3 a)$

$c^{2} - 2 a c - 2 b c + 4 a b = c^{2} - 3 a c - 3 b c + 9 a b$

$a c + b c = 5 a b$

$\frac{1}{a} + \frac{1}{b} = \frac{5}{c}$

Suggest Corrections

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