analytic geometry
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.
Let's consider an easier case, when the given lines have the form $y=mx$ and $y=nx$ (with $m\ne n$, of course).
The points on the angle bisectors satisfy the conditions that their distance from the two lines is the same. So you have, for such a point $(x,y)$, $$ \frac{|y-mx|}{\sqrt{1+m^2}}=\frac{|y-nx|}{\sqrt{1+n^2}} $$ By squaring, we get $$ (y-mx)^2(1+n^2)=(y-nx)^2(1+m^2) $$ and expanding $$ (y-mx)^2-(y-nx)^2=m^2(y-nx)^2-n^2(y-mx)^2. $$ The left hand side becomes $$ (y-mx+y-nx)(y-mx-y+nx)=-(m-n)(2y-(m+n)x)x $$ and the right hand side is $$ (my-mnx+ny-mnx)(my-mnx-ny+mnx)=(m-n)((m+n)y-2mnx)y $$ Since $m\ne n$, we can factor out $m-n$ on both sides, getting $$ -2xy+(m+n)x^2=(m+n)y^2-2mnxy $$ that can be written $$ (m+n)x^2-2(1-mn)xy-(m+n)y^2=0 $$ If you consider the equation $by^2+2hxy+ax^2=0$ in the unknown $y$, you can see that $m$ and $n$ are the roots of it (when $b\ne 0$). Thus $m+n=-2h/b$ and $mn=a/b$, so you can rewrite the above equation in the form $$ \frac{-2h}{b}x^2-2\left(1-\frac{a}{b}\right)xy-\frac{-2h}{b}y^2=0 $$ which is $$ hx^2-(a-b)xy-hy^2=0 $$ Note that it can't be both $h=0$ and $a=b\ne0$, because in this case the original equations would be $x^2+y^2=0$ (the isotropic lines, if you consider the complex plane).
$$x^2 - y^2 + 2y = 1\iff x^2 = y^2 -2y+1 = (y-1)^2 \iff y-1 = \pm x \iff y = 1\pm x$$
Thus the two intersecting lines are given by $y = 1+x$ and $y = 1-x$. The point of intersection is given by $(0, 1)$. One bisector is the y-axis: the line $x=0$. The other bisector is given by $y = 1$.
Can you take it from here?
Best Answer
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 .