Question

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

Answer 1

Let $A(x_1,y_1)$ and $B(x_2,y_2)$ be the given points.

Let O be the origin.

Slope of OA $=m_1=\frac{y_1}{x_1}$

Slope of OB $=m_2=\frac{y_2}{x_2}$

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 $cos\theta =\sqrt{\frac{1}{1+ta{n}^2\theta }}$

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

$\Rightarrow cos\theta =\frac{x_1{x}_2+y_1{y}_2}{\sqrt{x_2^2{y}_1^2+x_1^2{y}_2^2+x_1^2{x}_2^2+y_1^2{y}_2^2}}$

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