The coordinates of the orthocentre of the tri | vedclass

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(B) The given pair of lines is $2x^2 - 2y^2 + 3xy + 3x + y + 1 = 0$.Factorizing the expression: $2x^2 + 3xy - 2y^2 + 3x + y + 1 = 0$.$(x + 2y + 1)(2x

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$\left( -\frac{3}{5}, -\frac{1}{5} \right)$

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(B) The given pair of lines is $2x^2 - 2y^2 + 3xy + 3x + y + 1 = 0$.Factorizing the expression: $2x^2 + 3xy - 2y^2 + 3x + y + 1 = 0$.$(x + 2y + 1)(2x - y + 1) = 0$.Thus,the sides of the triangle are:$L_1: x + 2y + 1 = 0$$L_2: 2x - y + 1 = 0$$L_3: 3x + 2y + 1 = 0$Since the product of the slopes of $L_1$ and $L_2$ is $(-\frac{1}{2}) 	imes (2) = -1$,the lines $L_1$ and $L_2$ are perpendicular.In a right-angled triangle,the orthocentre is the vertex where the right angle is formed.Solving $x + 2y + 1 = 0$ and $2x - y + 1 = 0$:From $y = 2x + 1$,substitute into $x + 2(2x + 1) + 1 = 0$ $\Rightarrow 5x + 3 = 0$ $\Rightarrow x = -\frac{3}{5}$.Then $y = 2(-\frac{3}{5}) + 1 = -\frac{6}{5} + \frac{5}{5} = -\frac{1}{5}$.The orthocentre is $\left( -\frac{3}{5}, -\frac{1}{5} ight)$.

The coordinates of the orthocentre of the triangle formed by the lines $2x^2 - 2y^2 + 3xy + 3x + y + 1 = 0$ and $3x + 2y + 1 = 0$ are:

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