The orthocentre of the triangle formed by the | vedclass

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(A) Let the lines be $L_1: x + y - 1 = 0$,$L_2: 2x + 3y - 6 = 0$,and $L_3: 4x - y + 4 = 0$.To find the vertices,we solve the equations in pairs:$1$. $

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(A) Let the lines be $L_1: x + y - 1 = 0$,$L_2: 2x + 3y - 6 = 0$,and $L_3: 4x - y + 4 = 0$.To find the vertices,we solve the equations in pairs:$1$. $L_1$ and $L_2$: $x + y = 1$ and $2x + 3y = 6$. Solving gives $x = -3, y = 4$. Vertex $A = (-3, 4)$.$2$. $L_2$ and $L_3$: $2x + 3y = 6$ and $4x - y = -4$. Solving gives $x = -\frac{3}{7}, y = \frac{16}{7}$. Vertex $B = (-\frac{3}{7}, \frac{16}{7})$.$3$. $L_1$ and $L_3$: $x + y = 1$ and $4x - y = -4$. Solving gives $x = -\frac{3}{5}, y = \frac{8}{5}$. Vertex $C = (-\frac{3}{5}, \frac{8}{5})$.To find the orthocentre,we find the equations of altitudes:Altitude from $A$ to $BC$: Slope of $BC = \frac{8/5 - 16/7}{-3/5 - (-3/7)} = \frac{56-80}{35} / \frac{-21+15}{35} = \frac{-24}{-6} = 4$. The altitude is perpendicular to $BC$,so its slope is $-\frac{1}{4}$. Equation: $y - 4 = -\frac{1}{4}(x + 3) \implies x + 4y = 13$.Altitude from $B$ to $AC$: Slope of $AC = \frac{8/5 - 4}{-3/5 - (-3)} = \frac{-12/5}{12/5} = -1$. The altitude is perpendicular to $AC$,so its slope is $1$. Equation: $y - \frac{16}{7} = 1(x + \frac{3}{7}) \implies x - y = -\frac{19}{7} \implies 7x - 7y = -19$.Solving $x + 4y = 13$ and $7x - 7y = -19$: From the first,$x = 13 - 4y$. Substituting into the second: $7(13 - 4y) - 7y = -19 \implies 91 - 28y - 7y = -19 \implies -35y = -110 \implies y = \frac{22}{7}$. Then $x = 13 - 4(\frac{22}{7}) = \frac{91 - 88}{7} = \frac{3}{7}$.The orthocentre is $(\frac{3}{7}, \frac{22}{7})$,which lies in the First quadrant.

The orthocentre of the triangle formed by the lines $x + y = 1$,$2x + 3y = 6$,and $4x - y + 4 = 0$ lies in which quadrant?

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