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Find the centroid of the triangle formed by the pair of straight lines $12x^2-20xy+7y^2=0$ and the line $2x-3y+4=0$.

My doubt is:

The given pair of straight lines and the third line all pass through the point $(1,2)$. So how can three concurrent straight lines form a triangle? If the question has no flaw, please help me with it.

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1 Answer

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All you need to do is factorize the pair of equation of lines ie

$12x^2−20xy+7y^2=0$

$(6x-7y)(2x-y) = 0$

So these are two lines and $ (1,2)$ satisfies only one of them, not both of them . They form a triangle .

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So the centroid is $\left(\frac83,\frac83\right)$ –  Tejas Adsul Dec 5 '13 at 9:43
    
Is there a generalised method to find the equations of the two lines individually which form a pair of straight lines whose equation is given? –  Tejas Adsul Dec 5 '13 at 9:44
    
If they are a pair of straight lines then they will be factorisable . There are conditions to check whether the given quadratic in (x,y) is a pair of straight line or not . Refer to any book on co-ordinate geometry you will find it . –  AbKDs Dec 5 '13 at 9:47
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