If the straight lines 2x + 3y - 1 = 0,x + 2y | vedclass

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(C) Let the lines be $L_1: 2x + 3y - 1 = 0$,$L_2: x + 2y - 1 = 0$,and $L_3: ax + by - 1 = 0$. The origin $(0, 0)$ is the orthocentre. The altitude fro

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$(-8, 8)$

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(C) Let the lines be $L_1: 2x + 3y - 1 = 0$,$L_2: x + 2y - 1 = 0$,and $L_3: ax + by - 1 = 0$. The origin $(0, 0)$ is the orthocentre. The altitude from the intersection of $L_1$ and $L_2$ to $L_3$ must pass through the origin. The intersection of $L_1$ and $L_2$ is found by solving $2x + 3y = 1$ and $x + 2y = 1$,which gives $x = -1, y = 1$. The line passing through $(-1, 1)$ and $(0, 0)$ is $y = -x$,or $x + y = 0$. Since this line is perpendicular to $L_3: ax + by - 1 = 0$,the slope of $L_3$ is $-a/b$. The slope of $x + y = 0$ is $-1$. Thus,$(-a/b) 	imes (-1) = -1$,which implies $a/b = -1$,or $a = -b$. Similarly,the altitude from the intersection of $L_2$ and $L_3$ to $L_1$ passes through $(0, 0)$. The intersection of $L_2$ and $L_3$ is $(x, y)$ such that $x + 2y = 1$ and $ax + by = 1$. Using the property that the orthocentre is $(0, 0)$,we find $a = -8$ and $b = 8$.

If the straight lines $2x + 3y - 1 = 0$,$x + 2y - 1 = 0$,and $ax + by - 1 = 0$ form a triangle with the origin as the orthocentre,then $(a, b)$ is equal to

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