Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

up vote 1 down vote accepted

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


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

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

share|improve this answer
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 . –  Zoro Dec 5 '13 at 9:47

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.