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

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(A) Let the lines be $L_1: 2x + 3y - 1 = 0$,$L_2: x + 2y - 1 = 0$,and $L_3: ax + by - 1 = 0$. Since the orthocentre is at the origin $(0, 0)$,the alti

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$(-8, 8)$

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(A) Let the lines be $L_1: 2x + 3y - 1 = 0$,$L_2: x + 2y - 1 = 0$,and $L_3: ax + by - 1 = 0$. Since the orthocentre is at the origin $(0, 0)$,the altitude from the vertex formed by $L_1$ 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 $x = -1$ and $y = 1$. The altitude from $(-1, 1)$ to $L_3$ passes through $(0, 0)$,so its equation is $y = -x$,or $x + y = 0$. Since this altitude is perpendicular to $L_3: ax + by - 1 = 0$,the slope of $L_3$ must be $1$. Thus,$a/b = -1$,or $a = -b$. Substituting $b = -a$ into $L_3$,we get $ax - ay - 1 = 0$. Since the orthocentre is the intersection of altitudes,the altitude from the intersection of $L_2$ and $L_3$ to $L_1$ must also pass through the origin. The line perpendicular to $L_1$ passing through the origin is $3x - 2y = 0$. Solving $3x - 2y = 0$ and $x + 2y = 1$ gives $x = 1/4$ and $y = 3/8$. Substituting $(1/4, 3/8)$ into $ax - ay - 1 = 0$ gives $a(1/4 - 3/8) = 1$,so $a(-1/8) = 1$,which means $a = -8$. Since $b = -a$,we have $b = 8$. Thus,$(a, b) = (-8, 8)$.

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

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