The angle between the pair of straight lines | vedclass

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(A) The given equation is $x^2 - (y^2 + 2y + 1) = 0$.This can be written as $x^2 - (y + 1)^2 = 0$.Using the identity $a^2 - b^2 = (a - b)(a + b)$,we g

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$90$

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(A) The given equation is $x^2 - (y^2 + 2y + 1) = 0$.This can be written as $x^2 - (y + 1)^2 = 0$.Using the identity $a^2 - b^2 = (a - b)(a + b)$,we get $(x - (y + 1))(x + (y + 1)) = 0$.So,the two lines are $x - y - 1 = 0$ and $x + y + 1 = 0$.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.Alternatively,for a pair of lines $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,the lines are perpendicular if $a + b = 0$.Here,$a = 1$ and $b = -1$,so $a + b = 1 + (-1) = 0$.Therefore,the angle between the lines is $90^o$.

The angle between the pair of straight lines $x^2 - y^2 - 2y - 1 = 0$ is....$^o$

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