Question

If $\theta$ is the angle which the straight line joining the points $\left( {{x}_{1}},{{y}_{1}} \right)$and $\left( {{x}_{2}},{{y}_{2}} \right)$subtends at the origin prove that $\tan \theta =\dfrac{{{x}_{2}}{{y}_{1}}-{{x}_{1}}{{y}_{2}}}{{{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}}}$ and $\cos \theta =\dfrac{{{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}}}{\sqrt{{{x}_{1}}^{2}+{{y}_{1}}^{2}}\sqrt{{{x}_{2}}^{2}+{{y}_{2}}^{2}}}$.

Answer 1

Hint:Find the slope of the line formed by point A and B with origin because that will be the angle subtended by line AB on the origin , then use the Formula $\tan \theta =\left| \dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|$to find $\theta$ to find the angle between those two lines.

Complete step-by-step answer:
Let A $\left( {{x}_{1}},{{y}_{1}} \right)$ and B $\left( {{x}_{2}},{{y}_{2}} \right)$ be the given points.
Let O be the origin.
Slope of OA = ${{m}_{1}}=\dfrac{{{y}_{1}}}{{{x}_{1}}}$
Slope of OB = ${{m}_{2}}=\dfrac{{{y}_{2}}}{{{x}_{2}}}$
It is given that $\theta$ is the angle between lines OA and OB.
$\tan \theta =\left| \dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|$
By substituting the values of A and B we get,
$\tan \theta =\dfrac{\dfrac{{{y}_{1}}}{{{x}_{1}}}-\dfrac{{{y}_{2}}}{{{x}_{2}}}}{1+\left( \dfrac{{{y}_{1}}}{{{x}_{1}}}\times \dfrac{{{y}_{2}}}{{{x}_{2}}} \right)}$
By solving this we get,
$\tan \theta =\dfrac{{{x}_{2}}{{y}_{1}}-{{x}_{1}}{{y}_{2}}}{{{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}}}$ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
We know that $\cos \theta =\sqrt{\dfrac{1}{1+{{\tan }^{2}}\theta }}$. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)
Substitute the value (1) in (2), we get
$\cos \theta =\dfrac{{{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}}}{\sqrt{{{({{x}_{2}}{{y}_{1}}-{{x}_{1}}{{y}_{2}})}^{2}}+{{\left( {{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}} \right)}^{2}}}}$
$\cos \theta =\dfrac{{{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}}}{\sqrt{{{x}_{2}}^{2}{{y}_{1}}^{2}+{{x}_{1}}^{2}{{y}_{2}}^{2}+{{x}_{1}}^{2}{{x}_{2}}^{2}+{{y}_{1}}^{2}{{y}_{2}}^{2}}}$
$\cos \theta =\dfrac{{{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}}}{\sqrt{{{x}_{1}}^{2}+{{y}_{1}}^{2}}\sqrt{{{x}_{2}}^{2}+{{y}_{2}}^{2}}}$
Hence proved.

Note: From the above we can conclude that after finding the slopes we get the value $\tan \theta$ and further substituted in the trigonometric operation to get the final solution. Only variables are used so modulus doesn’t play an important role.