# Question about a pair of straight lines

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.

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 .

• So the centroid is $\left(\frac83,\frac83\right)$ – Tejas 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 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

Actually if the point is satisfied by the given pair of straight lines, then the point need not lie on both the lines i.e.,$6x-7y=0$ and $2x-y=0$..

It may lie on any one of it. Here it lies only on $2x-y=0$ so they are not concurrent