The acute angle between two straight lines pa | vedclass

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Question

(B) The line $2x + y + 10 = 0$ intersects the $x$-axis at $A(-5, 0)$ and the $y$-axis at $B(0, -10)$.Let the points dividing $AB$ in the ratio $1:2:2$

Vedclass · Accepted Answer

$\pi /4$

Vedclass · Answer

(B) The line $2x + y + 10 = 0$ intersects the $x$-axis at $A(-5, 0)$ and the $y$-axis at $B(0, -10)$.Let the points dividing $AB$ in the ratio $1:2:2$ be $P$ and $Q$. The total ratio is $1+2+2 = 5$.The coordinates of $P$ are $\left( \frac{2(-5) + 3(0)}{5}, \frac{2(0) + 3(-10)}{5} ight) = (-2, -6)$.The coordinates of $Q$ are $\left( \frac{4(-5) + 1(0)}{5}, \frac{4(0) + 1(-10)}{5} ight) = (-4, -2)$.The slopes of lines $MP$ and $MQ$ are $m_1 = \frac{-6 - (-8)}{-2 - (-6)} = \frac{2}{4} = \frac{1}{2}$ and $m_2 = \frac{-2 - (-8)}{-4 - (-6)} = \frac{6}{2} = 3$.The angle $	heta$ between the lines is given by $	an 	heta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} ight| = \left| \frac{3 - 0.5}{1 + 3(0.5)} ight| = \left| \frac{2.5}{2.5} ight| = 1$.Thus,$	heta = 	an^{-1}(1) = \pi / 4$.

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