Question

If θ is the angle which the straight line joining the points (x₁, y₁) and (x₂, y₂) subtends at the origin, prove that  \[\tan \theta = \frac{x_2 y_1 - x_1 y_2}{x_1 x_2 + y_1 y_2}\text { and } \cos \theta = \frac{x_1 x_2 + y_1 y_2}{\sqrt{{x_1}^2 + {y_1}^2}\sqrt{{x_2}^2 + {y_2}^2}}\].

Answer 1

Let A (x₁, y₁) and B (x₂, y₂) be the given points.
Let O be the origin.

Slope of OA = m₁ = \[\frac{y_1}{x_1}\]

Slope of OB = m₂ = \[\frac{y_2}{x_2}\]

It is given that \[\theta\] is the angle between lines OA and OB.

\[\therefore \tan \theta = \frac{m_1 - m_2}{1 + m_1 m_2}\]

\[ = \frac{\frac{y_1}{x_1} - \frac{y_2}{x_2}}{1 + \frac{y_1}{x_1} \times \frac{y_2}{x_2}}\]

\[ \Rightarrow \tan \theta = \frac{x_2 y_1 - x_1 y_2}{x_1 x_2 + y_1 y_2}\]

Now,
As we know that

\[\cos \theta = \sqrt{\frac{1}{1 + \tan^2 \theta}}\]

\[\therefore \cos \theta = \frac{x_1 x_2 + y_1 y_2}{\sqrt{\left( x_2 y_1 - x_1 y_2 \right)^2 + \left( x_1 x_2 + y_1 y_2 \right)^2}}\]

\[\Rightarrow \cos \theta = \frac{x_1 x_2 + y_1 y_2}{\sqrt{{x_2}^2 {y_1}^2 + {x_1}^2 {y_2}^2 + {x_1}^2 {x_2}^2 + {y_1}^2 {y_2}^2}}\]

\[\Rightarrow \cos \theta = \frac{x_1 x_2 + y_1 y_2}{\sqrt{{x_1}^2 + {y_1}^2} \sqrt{{x_2}^2 + {y_2}^2}}\]