If left(frac{2}{3}, 0right) is the centroid o | vedclass

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(B) The equation $4x^2-y^2=0$ can be factored as $(2x-y)(2x+y)=0$,which represents two lines: $L_1: 2x-y=0$ and $L_2: 2x+y=0$.Let the third line be $L

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(B) The equation $4x^2-y^2=0$ can be factored as $(2x-y)(2x+y)=0$,which represents two lines: $L_1: 2x-y=0$ and $L_2: 2x+y=0$.Let the third line be $L_3: lx+my+n=0$.The vertices of the triangle are the intersection points of these lines.Intersection of $L_1$ and $L_2$: $(0, 0)$.Intersection of $L_1$ and $L_3$: $2x-y=0 \implies y=2x$. Substituting into $L_3$: $lx+m(2x)+n=0 \implies x_1 = -\frac{n}{l+2m}, y_1 = -\frac{2n}{l+2m}$.Intersection of $L_2$ and $L_3$: $2x+y=0 \implies y=-2x$. Substituting into $L_3$: $lx+m(-2x)+n=0 \implies x_2 = -\frac{n}{l-2m}, y_2 = \frac{2n}{l-2m}$.The centroid $(G_x, G_y)$ is given by $\left(\frac{0+x_1+x_2}{3}, \frac{0+y_1+y_2}{3}ight) = \left(\frac{2}{3}, 0ight)$.Thus,$x_1+x_2 = 2$ and $y_1+y_2 = 0$.$y_1+y_2 = -2n \left(\frac{1}{l+2m} - \frac{1}{l-2m}ight) = -2n \left(\frac{l-2m-l-2m}{l^2-4m^2}ight) = \frac{8mn}{l^2-4m^2} = 0$. Since $n 
eq 0$,we must have $m=0$.Then $x_1+x_2 = -\frac{n}{l} - \frac{n}{l} = -\frac{2n}{l} = 2 \implies l = -n$.Since $m=0$ and $l=-n$,the line $lx+my+n=0$ becomes $-nx+n=0$,which simplifies to $x=1$.For the centroid to be $\left(\frac{2}{3}, 0ight)$,the line $lx+my+n=0$ must be $x=1$,so $l=1, m=0, n=-1$.Then $l+m+n = 1+0-1 = 0$.

$X$	$1$	$2$	$3$	$4$	$5$
$P(X)$	$k^2$	$2k$	$k$	$2k$	$5k^2$

If $\left(\frac{2}{3}, 0\right)$ is the centroid of the triangle formed by the lines $4x^2-y^2=0$ and $lx+my+n=0$,then $l+m+n=$

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