The inclination of the straight line passing | vedclass

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Question

(D) First,find the midpoint of the line segment joining $(4, -5)$ and $(-2, 9)$.Midpoint $= \left( \frac{4 + (-2)}{2}, \frac{-5 + 9}{2} \right) = \lef

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$3\pi /4$

Vedclass · Answer

(D) First,find the midpoint of the line segment joining $(4, -5)$ and $(-2, 9)$.Midpoint $= \left( \frac{4 + (-2)}{2}, \frac{-5 + 9}{2} ight) = \left( \frac{2}{2}, \frac{4}{2} ight) = (1, 2)$.Next,find the slope $m$ of the line passing through $(-3, 6)$ and $(1, 2)$.$m = \frac{2 - 6}{1 - (-3)} = \frac{-4}{4} = -1$.Since $m = 	an 	heta$,we have $	an 	heta = -1$.Since $	an 	heta = -1$,the inclination $	heta$ is $\pi - \frac{\pi}{4} = \frac{3\pi}{4}$.

The inclination of the straight line passing through the point $(-3, 6)$ and the midpoint of the line joining the points $(4, -5)$ and $(-2, 9)$ is

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