The lines represented by the equation x^2-y^2 | vedclass

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(D) The given equation is $x^2-y^2-x+3y-2=0$. We can factorize this expression by grouping terms: $x^2 - (y^2 - 3y + 2) = 0$ $x^2 - (y-1)(y-2) = 0$ Th

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$x-y+1=0, x+y-2=0$

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(D) The given equation is $x^2-y^2-x+3y-2=0$. We can factorize this expression by grouping terms: $x^2 - (y^2 - 3y + 2) = 0$ $x^2 - (y-1)(y-2) = 0$ This does not immediately factor into linear factors. Let us rewrite the equation as $x^2 - x - (y^2 - 3y + 2) = 0$. Using the quadratic formula for $x$ in terms of $y$: $x = \frac{1 \pm \sqrt{1 + 4(y^2 - 3y + 2)}}{2} = \frac{1 \pm \sqrt{4y^2 - 12y + 9}}{2} = \frac{1 \pm \sqrt{(2y-3)^2}}{2} = \frac{1 \pm (2y-3)}{2}$. Case $1$: $x = \frac{1 + 2y - 3}{2} = \frac{2y-2}{2} = y-1 \implies x-y+1=0$. Case $2$: $x = \frac{1 - 2y + 3}{2} = \frac{4-2y}{2} = 2-y \implies x+y-2=0$. Thus,the lines are $x-y+1=0$ and $x+y-2=0$.

The lines represented by the equation $x^2-y^2-x+3y-2=0$ are :

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