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VITEEE 2010 Paper

Question : 120 of 120

Marks: +1, -0

Solution:

A (-2,1), B (2,3) and C (-2,-4) are three given points.

Slope of the line BA

m_{1}=\frac{1-3}{-2-2}=\frac{1}{2}

\left[\text{Using slope formula, } m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \right]

Slope of the line BC

m_{2}=\frac{-4-3}{-2-2}=\frac{7}{4}

Now, angle between AB and BC is given by

\tan\theta = \left| \frac{m_{1}-m_{2}}{1+m_{1}m_{2}} \right| = \left| \frac{\frac{1}{2}-\frac{7}{4}}{1+\frac{1}{2}\cdot\frac{7}{4}} \right|

⇒

\tan\theta = \left| -\frac{10}{15} \right|

⇒

\tan\theta = \left| -\frac{2}{3} \right|

⇒

\theta = \tan^{-1}\left| -\frac{2}{3} \right|

⇒

\theta = \tan^{-1}\left( \frac{2}{3} \right) \;\;\;\;\;\;\;\; [\because |-x| = x]

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