Question:

Find perpendicular distance of the line joining the points $(cos \,\theta, sin \,\theta)$ and $(cos\, \phi, sin \,\phi)$ from the origin.

Updated On: Jul 6, 2022
  • $\left|cos\left(\frac{\theta-\phi}{2}\right)\right|$
  • $\left|cos\left(\frac{\theta+\phi}{2}\right)\right|$
  • $\left|sin\left(\frac{\theta+\phi}{2}\right)\right|$
  • None of these
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The Correct Option is A

Solution and Explanation

Let the points be $A = \left(cos\, \theta, sin\, \theta\right)$ and $B = \left(cos\,\phi, sin\,\phi\right)$. Equation of line $AB$ is $y-sin\,\theta=\frac{sin\, \phi-sin\,\theta}{cos\,\phi-cos\,\theta}\left(x-cos\,\phi\right)$ or $y-sin\,\theta=\frac{2\,cos\left(\frac{\theta+\phi}{2}\right)sin\left(\frac{\phi-\theta}{2}\right)}{-2\, sin\left(\frac{\phi+\theta}{2}\right)sin\left(\frac{\phi-\theta}{2}\right)}\left(x-cos\, \theta\right)$ or $y\, sin\left(\frac{\theta+\phi}{2}\right)-sin\, \theta\,sin\left(\frac{\theta+\phi}{2}\right)$ $=-x\,cos\left(\frac{\theta+\phi}{2}\right)+cos\,\theta\,cos\left(\frac{\theta+\phi}{2}\right)$ or $x\,cos\left(\frac{\theta +\phi }{2}\right)+y\,sin\left(\frac{\theta +\phi }{2}\right)$ $-\left\{cos\,\theta\,cos\left(\frac{\theta +\phi }{2}\right)+sin\,\theta\,sin\left(\frac{\theta +\phi }{2}\right)\right\}=0$ or $x\,cos\left(\frac{\theta +\phi }{2}\right)+y\,sin\left(\frac{\theta +\phi }{2}\right)-cos\left(\theta-\frac{\theta +\phi }{2}\right)=0$ or $x\,cos\left(\frac{\theta +\phi }{2}\right)+y\,sin\left(\frac{\theta +\phi }{2}\right)-cos\left(\frac{\theta -\phi }{2}\right)=0$ $\therefore$ Perpendicular distance of line from the origin $=\frac{\left|0+0-cos\left(\frac{\theta-\phi}{2}\right)\right|}{\sqrt{cos^{2}\left(\frac{\theta+\phi}{2}\right)+sin^{2}\left(\frac{\theta+\phi}{2}\right)}}$ $=\left|cos\left(\frac{\theta-\phi}{2}\right)\right|$
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Top Questions on Straight lines

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Notes on Straight lines

Concepts Used:

Straight lines

A straight line is a line having the shortest distance between two points. 

A straight line can be represented as an equation in various forms,  as show in the image below:

 

The following are the many forms of the equation of the line that are presented in straight line-

1. Slope – Point Form

Assume P0(x0, y0) is a fixed point on a non-vertical line L with m as its slope. If P (x, y) is an arbitrary point on L, then the point (x, y) lies on the line with slope m through the fixed point (x0, y0) if and only if its coordinates fulfil the equation below.

y – y0 = m (x – x0)

2. Two – Point Form

Let's look at the line. L crosses between two places. P1(x1, y1) and P2(x2, y2)  are general points on L, while P (x, y) is a general point on L. As a result, the three points P1, P2, and P are collinear, and it becomes

The slope of P2P = The slope of P1P2 , i.e.

\(\frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1}\)

Hence, the equation becomes:

y - y1 =\( \frac{y_2-y_1}{x_2-x_1} (x-x1)\)

3. Slope-Intercept Form

Assume that a line L with slope m intersects the y-axis at a distance c from the origin, and that the distance c is referred to as the line L's y-intercept. As a result, the coordinates of the spot on the y-axis where the line intersects are (0, c). As a result, the slope of the line L is m, and it passes through a fixed point (0, c). The equation of the line L thus obtained from the slope – point form is given by

y – c =m( x - 0 )

As a result, the point (x, y) on the line with slope m and y-intercept c lies on the line, if and only if

y = m x +c