0
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 .

$\endgroup$
  • $\begingroup$ So the centroid is $\left(\frac83,\frac83\right)$ $\endgroup$ – Tejas Dec 5 '13 at 9:43
  • $\begingroup$ 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? $\endgroup$ – Tejas Dec 5 '13 at 9:44
  • $\begingroup$ 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 . $\endgroup$ – AbKDs Dec 5 '13 at 9:47
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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