Question

The area of the triangle formed by the line $\mathrm{x}\sin \mathrm{\alpha }+\mathrm{y}\cos \mathrm{\alpha }=\sin 2\mathrm{\alpha }$ and the coordinates axes is

• A. $\sin 2\mathrm{\alpha }$
• B. $\cos 2\mathrm{\alpha }$
• C. $2\sin 2\mathrm{\alpha }$
• D. $2\cos 2\mathrm{\alpha }$

Answer 1

The correct option is A

$\sin 2\mathrm{\alpha }$

Explanation for the correct option.

Determine the area of the triangle and coordinate axes of the formed triangle.

We have the line equation as: $\mathrm{xsin\alpha }+\mathrm{ycos\alpha }=\sin 2\mathrm{\alpha }$

Putting the value of $\mathrm{x}=0\mathrm{and}\mathrm{y}=0$, to get the coordinates of the point of intersection.

When $\mathrm{x}=0,$

$$\begin{gathered} \Rightarrow \mathrm{y}\cos \mathrm{\alpha }=\sin 2\alpha \\ \Rightarrow \mathrm{y}\cos \mathrm{\alpha }=2\times \sin \mathrm{\alpha }\times \cos \mathrm{\alpha }[\because \sin 2\mathrm{\alpha }=2\sin \mathrm{\alpha }\times \cos \mathrm{\alpha }] \\ \Rightarrow \mathrm{y}=2\sin \mathrm{\alpha } \end{gathered}$$

Similarly, when $y=0,$

$$\begin{gathered} \Rightarrow x\sin \mathrm{\alpha }=\sin 2\alpha \\ \Rightarrow x\sin \mathrm{\alpha }=2\times \sin \mathrm{\alpha }\times \cos \mathrm{\alpha }[\because \sin 2\mathrm{\alpha }=2\sin \mathrm{\alpha }\times \cos \mathrm{\alpha }] \\ \Rightarrow \mathrm{y}=2\cos \mathrm{\alpha } \end{gathered}$$

Applying area of triangle formula:

$\text{Area}=\frac{1}{2}\times \text{base×height}$

$$\begin{gathered} =\frac{1}{2}\times 2\mathrm{cos\alpha }\times 2\mathrm{sin\alpha } \\ =2\times \sin \alpha \times \cos \alpha \\ =\sin 2\mathrm{\alpha }[\because \sin 2\mathrm{\alpha }=2\times \sin \mathrm{\alpha }\times \cos \mathrm{\alpha }] \end{gathered}$$

Hence, option (A) is correct.