Consider the lines L_1 equiv 4x + 5y - 6 = 0, | vedclass

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(A) Given lines are:$L_1 \equiv 4x + 5y - 6 = 0 \dots(1)$$L_2 \equiv 2x + 3y - 4 = 0 \dots(2)$$L_3 \equiv 3x - y + 2 = 0 \dots(3)$Point $A$ is the int

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$26x^2 + 17xy + 2y^2 = 0$

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(A) Given lines are:$L_1 \equiv 4x + 5y - 6 = 0 \dots(1)$$L_2 \equiv 2x + 3y - 4 = 0 \dots(2)$$L_3 \equiv 3x - y + 2 = 0 \dots(3)$Point $A$ is the intersection of $L_1$ and $L_2$. Solving $(1)$ and $(2)$:$4x + 5y = 6$$4x + 6y = 8$Subtracting gives $y = 2$,then $x = -1$. So,$A = (-1, 2)$.Point $B$ is the intersection of $L_1$ and $L_3$. Solving $(1)$ and $(3)$:$4x + 5y = 6$$15x - 5y = -10$Adding gives $19x = -4$,so $x = -4/19$. Then $y = 3x + 2 = 3(-4/19) + 2 = (-12 + 38)/19 = 26/19$. So,$B = (-4/19, 26/19)$.The equation of line $OA$ (passing through $(0,0)$ and $(-1,2)$) is $y = -2x$,or $2x + y = 0$.The equation of line $OB$ (passing through $(0,0)$ and $(-4/19, 26/19)$) is $y = (26/19) / (-4/19) x = -13/2 x$,or $13x + 2y = 0$.The combined equation of $OA$ and $OB$ is $(2x + y)(13x + 2y) = 0$.Expanding this: $26x^2 + 4xy + 13xy + 2y^2 = 0$,which simplifies to $26x^2 + 17xy + 2y^2 = 0$.

Consider the lines $L_1 \equiv 4x + 5y - 6 = 0$,$L_2 \equiv 2x + 3y - 4 = 0$,and $L_3 \equiv 3x - y + 2 = 0$. If the line $L_1 = 0$ intersects the lines $L_2 = 0$ and $L_3 = 0$ at the points $A$ and $B$ respectively,then the combined equation of lines $OA$ and $OB$ is

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