The angle between the lines represented by x^ | vedclass

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(A) The given equation is $x^2 - y^2 - 2y - 1 = 0$. This can be rewritten as $x^2 - (y^2 + 2y + 1) = 0$. This simplifies to $x^2 - (y + 1)^2 = 0$. Usi

Vedclass · Accepted Answer

$90$

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(A) The given equation is $x^2 - y^2 - 2y - 1 = 0$. This can be rewritten as $x^2 - (y^2 + 2y + 1) = 0$. This simplifies to $x^2 - (y + 1)^2 = 0$. Using the difference of squares formula $a^2 - b^2 = (a - b)(a + b)$,we get: $(x - (y + 1))(x + (y + 1)) = 0$. $(x - y - 1)(x + y + 1) = 0$. Thus,the two lines are $L_1: x - y - 1 = 0$ and $L_2: x + y + 1 = 0$. The slopes of these lines are $m_1 = 1$ and $m_2 = -1$. Since $m_1 	imes m_2 = 1 	imes (-1) = -1$,the lines are perpendicular to each other. Therefore,the angle between them is $90^o$.

The angle between the lines represented by $x^2 - y^2 - 2y - 1 = 0$ is .....$^o$

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