To find the locus of midpoints, first find the general coordinates of the midpoint in terms of a parameter (like \( \theta \) in this case). Then, eliminate the parameter to get the equation of the locus.
The given equation of the line is:
\( x\cos\theta + y\sin\theta = 7 \)
Step 1: Intercepts of the line
The x-intercept is: \( \frac{7}{\cos\theta} \)
The y-intercept is: \( \frac{7}{\sin\theta} \)
Thus, the intercept points are:
\( A \left( \frac{7}{\cos\theta} , 0 \right) \) and \( B \left( 0, \frac{7}{\sin\theta} \right) \)
Step 2: Coordinates of the midpoint \(M(h,k)\)
The midpoint of \( A \) and \( B \) is given by:
\( h = \frac{7}{2\cos\theta}, \quad k = \frac{7}{2\sin\theta} \)
Step 3: Solving for \( \theta \)
From the given information:
\( k = \frac{7\sqrt{3}}{3} \)
Substitute \( k = \frac{7}{2\sin\theta} \):
\( \frac{7}{2\sin\theta} = \frac{7\sqrt{3}}{3} \Rightarrow \sin\theta = \frac{\sqrt{3}}{2} \)
Thus:
\( \theta = \frac{\pi}{3} \)
Step 4: Solving for \( \alpha \)
Using \( h = \frac{7}{2\cos\theta} \):
\( \alpha = \frac{7}{2\cos\theta} = \frac{7}{2\cos(\frac{\pi}{3})} = \frac{7}{2(\frac{1}{2})} = 7 \)
The value of current \( I \) in the electrical circuit as given below, when the potential at \( A \) is equal to the potential at \( B \), will be _____ A.
Two light beams fall on a transparent material block at point 1 and 2 with angle \( \theta_1 \) and \( \theta_2 \), respectively, as shown in the figure. After refraction, the beams intersect at point 3 which is exactly on the interface at the other end of the block. Given: the distance between 1 and 2, \( d = \frac{4}{3} \) cm and \( \theta_1 = \theta_2 = \cos^{-1} \left( \frac{n_2}{2n_1} \right) \), where \( n_2 \) is the refractive index of the block and \( n_1 \) is the refractive index of the outside medium, then the thickness of the block is …….. cm.