The angle between the line joining the points | vedclass

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(B) Let the line joining the points $(1, -2)$ and $(3, 2)$ be $L_1$. The slope $m_1$ of $L_1$ is given by $m_1 = \frac{2 - (-2)}{3 - 1} = \frac{4}{2}

Vedclass · Accepted Answer

$\pi / 2$

Vedclass · Answer

(B) Let the line joining the points $(1, -2)$ and $(3, 2)$ be $L_1$. The slope $m_1$ of $L_1$ is given by $m_1 = \frac{2 - (-2)}{3 - 1} = \frac{4}{2} = 2$.The equation of the second line $L_2$ is $x + 2y - 7 = 0$,which can be written as $2y = -x + 7$ or $y = -\frac{1}{2}x + \frac{7}{2}$.The slope $m_2$ of $L_2$ is $-\frac{1}{2}$.Now,calculate the product of the slopes: $m_1 	imes m_2 = 2 	imes (-\frac{1}{2}) = -1$.Since the product of the slopes is $-1$,the two lines are perpendicular to each other.Therefore,the angle between the lines is $\frac{\pi}{2}$.

The angle between the line joining the points $(1, -2)$ and $(3, 2)$ and the line $x + 2y - 7 = 0$ is

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