Question

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The distance from the origin to the line x cos theta + y sin theta - tan theta = 0 is​

Answer 1

★ Concept :-

Here the concept of Coordinate Geometry and Lines has been used. We see that we have to find the distance of the given line from the origin. Since here we aren't given any distance or any relationship of the distance, thus we shall find the shortest possible distance. This shortest possible distance is the perpendicular distance from origin to the line. We already have the formula for this. Firstly we shall get values from the given things and then apply it in the Formula to find the answer.

Let's do it !!

Let's do it !!_____________________________________

★ Formula Used :-

$\;\;\boxed{\sf{\pink{d\;=\;\dfrac{\bigg|Ax_{1}\;+\;By_{1}\;+\;C\bigg|}{\sqrt{A^{2}\:+\:B^{2}}}}}}$

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★ Solution :-

Given,

» Equation of Line :: x cos θ + y sin θ - tan θ = 0

» Coordinates of another point = (0, 0)

Here coordinates can be taken as (x₁, y₁)

where,

• x₁ = 0

• y₁ = 0

From the equation of line we have,

• A = cos θ

• B = sin θ

• C = tan θ

• Let the distance between the line and point be d

Now we already have the formula to calculate perpendicular distance between a line and another point. So,

$\;\;\tt{\rightarrow\;\;d\;=\;\dfrac{\bigg|Ax_{1}\;+\;By_{1}\;+\;C\bigg|}{\sqrt{A^{2}\:+\:B^{2}}}}$

• Here d is the perpendicular distance .

By applying the values of (x₁, y₁), we get

$\;\;\tt{\rightarrow\;\;d\;=\;\dfrac{\bigg|A(0)\;+\;B(0)\;+\;C\bigg|}{\sqrt{A^{2}\:+\:B^{2}}}}$

Now let's apply the value of A, B and C. Then,

$\;\;\tt{\rightarrow\;\;d\;=\;\dfrac{\bigg|(\cos\theta)(0)\;+\;(\sin\theta)(0)\;+\;(\tan\theta)\bigg|}{\sqrt{(\cos\theta)^{2}\:+\:(\sin\theta)^{2}}}}$

$\;\;\tt{\rightarrow\;\;d\;=\;\dfrac{\bigg|(0)\;+\;(0)\;+\;(\tan\theta)\bigg|}{\sqrt{\cos^{2}\theta\:+\:\sin^{2}\theta}}}$

$\;\;\tt{\rightarrow\;\;d\;=\;\dfrac{\bigg|(\tan\theta)\bigg|}{\sqrt{\cos^{2}\theta\:+\:\sin^{2}\theta}}}$

We know that,

• sin² p + cos² p = 1 (where p is an angle)

• Here p = θ

By applying the relationship, we get

$\;\;\tt{\rightarrow\;\;d\;=\;\dfrac{\bigg|(\tan\theta)\bigg|}{\sqrt{1}}}$

$\;\;\tt{\rightarrow\;\;d\;=\;\dfrac{\bigg|(\tan\theta)\bigg|}{\sqrt{1^{2}}}}$

• Since 1² = 1

$\;\;\tt{\rightarrow\;\;d\;=\;\dfrac{\bigg|(\tan\theta)\bigg|}{\pm\;1}}$

• Since, √x² = ± x

Now here we have two values of d.

$\;\;\tt{\rightarrow\;\;d\;=\;\dfrac{\bigg|(\tan\theta)\bigg|}{+\;1},\;\dfrac{\bigg|(\tan\theta)\bigg|}{-1}}$

$\;\;\tt{\rightarrow\;\;d\;=\;\dfrac{\bigg|(\tan\theta)\bigg|}{1},\;\dfrac{-\bigg[\bigg|(\tan\theta)\bigg|\bigg]}{1}}$

• Since distance is never taken as negative. Thus we have to neglect the negative value of d. So,

$\;\;\tt{\rightarrow\;\;d\;=\;\dfrac{\bigg|(\tan\theta)\bigg|}{1}}$

$\;\;\tt{\rightarrow\;\;d\;=\;\bigg|(\tan\theta)\bigg|}$

$\;\;\tt{\rightarrow\;\;d\;=\;(\tan\theta)}$

$\;\;\bf{\rightarrow\;\;\red{d\;=\;\tan\theta}}$

This is the required answer.

$\;\underline{\boxed{\tt{Required\;distance\;between\;the\;line\;and\;origin\;=\;\bf{\purple{\tan\theta}}}}}$

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★ More to know :-

• Distance Formula :: Distance between two points A and B is given as,

$\;\sf{\leadsto\;\;AB\;=\;\sqrt{(x_{2}\:-\:x_{1})^{2}\:+\:(y_{2}\:-\:y_{1})^{2}}}$

• Where x and y (along with subscripts) are the coordinates of two points.

• Slope of a Line :: Slope of a line (m) is given as,

$\;\sf{\leadsto\;\;m\;=\;\dfrac{y_{2}\:-\:y_{1}}{x_{2}\:-\:x_{1}},\;\;\quad x_{1}\:\neq\:x_{2}}$

• Equation of line with intercepts a and b on x - axis and y - axis ::

$\;\sf{\leadsto\;\;\dfrac{x}{a}\;+\;\dfrac{y}{b}\;=\;1}$

• Slope of line m and intercepts y axis a# c ::

>> y = mx + c

• Slope of line m and intercepts x axis as d ::

>> y = m(x - d)