triangle OAB is formed by the lines x^2-4xy+y | vedclass

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(B) Let $D$ be the midpoint of line $AB$. Let $A = (x_1, y_1)$ and $B = (x_2, y_2)$. Then $D = (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$. The combined e

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$7x-8y=0$

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(B) Let $D$ be the midpoint of line $AB$. Let $A = (x_1, y_1)$ and $B = (x_2, y_2)$. Then $D = (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$. The combined equation of sides $OA$ and $OB$ is $x^2-4xy+y^2=0$. The equation of line $AB$ is $2x+3y-1=0$,so $x = \frac{1-3y}{2}$. Substituting this into the combined equation: $(\frac{1-3y}{2})^2 - 4(\frac{1-3y}{2})y + y^2 = 0$ $(1-3y)^2 - 8y(1-3y) + 4y^2 = 0$ $1 - 6y + 9y^2 - 8y + 24y^2 + 4y^2 = 0$ $37y^2 - 14y + 1 = 0$. Sum of roots $y_1+y_2 = \frac{14}{37}$. $y$-coordinate of $D = \frac{y_1+y_2}{2} = \frac{7}{37}$. Since $D$ lies on $2x+3y-1=0$: $2x + 3(\frac{7}{37}) - 1 = 0$ $2x + \frac{21}{37} - 1 = 0$ $2x = 1 - \frac{21}{37} = \frac{16}{37} \Rightarrow x = \frac{8}{37}$. Thus,$D = (\frac{8}{37}, \frac{7}{37})$. The equation of the median $OD$ passing through $(0,0)$ and $(\frac{8}{37}, \frac{7}{37})$ is: $\frac{y-0}{x-0} = \frac{7/37}{8/37} = \frac{7}{8}$ $8y = 7x \Rightarrow 7x-8y=0$.

$\triangle OAB$ is formed by the lines $x^2-4xy+y^2=0$ and the line $AB$. The equation of line $AB$ is $2x+3y-1=0$. Then the equation of the median of the triangle drawn from the origin is

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