Question

If θ is the angle which the straight line joining the points (x₁, y₁) and (x₂, y₂) subtends at the origin, prove that
$\tan \mathrm{\theta }=\frac{x_2{y}_1-x_1{y}_2}{x_1{x}_2+y_1{y}_2}\mathrm{and}\cos \mathrm{\theta }=\frac{x_1{x}_2+y_1{y}_2}{\sqrt{{x_1}^2+{y_1}^2}\sqrt{{x_2}^2+{y_2}^2}}$

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.

$$\begin{gathered} \therefore \tan \theta =\frac{m_1-m_2}{1+m_1{m}_2} \\ =\frac{{\displaystyle \frac{y_1}{x_1}}-{\displaystyle \frac{y_2}{x_2}}}{1+{\displaystyle \frac{y_1}{x_1}\times \frac{y_2}{x_2}}} \\ \Rightarrow \tan \theta =\frac{{\displaystyle x_2{y}_1-x_1{y}_2}}{{\displaystyle x_1{x}_2+y_1{y}_2}} \end{gathered}$$


Now,

As we know that $\cos \theta =\sqrt{\frac{1}{1+{\tan }^2\theta }}$

$\therefore \cos \theta =\frac{x_1{x}_2+y_1{y}_2}{\sqrt{{\left(x_2{y}_1-x_1{y}_2\right)}^2+{\left(x_1{x}_2+y_1{y}_2\right)}^2}}$

$\Rightarrow \cos \theta =\frac{x_1{x}_2+y_1{y}_2}{\sqrt{{x_2}^2{y_1}^2+{x_1}^2{y_2}^2+{x_1}^2{x_2}^2+{y_1}^2{y_2}^2}}$

$\Rightarrow \cos \theta =\frac{x_1{x}_2+y_1{y}_2}{\sqrt{{x_1}^2+{y_1}^2}\sqrt{{x_2}^2+{y_2}^2}}$