Question

QUESTION :-
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☆ Show that the equation of the line passing through the origin and making an angle θ with the line y = mx + c is
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$\: \: \: \: \: \: \: \: \bullet\bf\: \: \: {\dfrac{y}{x} = \dfrac{m \pm tan \theta}{1 \mp m\: tan \theta}}$
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Answer 1

Given :

A line AB with equation y= mx+c

and another angle PQ who makes angle θ with AB

To prove that:

The equation of line PQ is :

$\frac yx=\frac{m\pm tan \theta}{ 1 \mp tan \theta}$

Solution :

The equation of AB is

y = mx+C-------------------------------(1)

so its slope m₁ = m

Let the equation of PQ is :

$y = m_1x + c_1$

This passes through origin

so 0=0+c₁

So c₁=0

Thus the equation of PQ becomes

$y = m_1x -----------------------(2)$

Now θ is angle between PQ and AB

$So, tan\, \theta = \pm{ \frac { m-m_1}{1-mm_1}}\\\\m-m_1 = \pm tan \theta \mp mm_1tan\theta\\\\We\ need \,to\,find\ value\ of \,m_1 ,So\ taking\, m_1\ in\ LHS\\\\\pm mm_1tan\theta-m_1 =-m\pm tan\theta\\\\m_1( 1 \mp \,tan\theta\,)=m\pm tan\theta\\\\m_1 = \frac{m\pm tan\theta}{ 1 - \mp \,tan\theta\,}$

On plugging value of m₁ in (2) we get the equation of PQ:

$y = \frac{m\pm tan\theta}{ 1 - \mp \,tan\theta\,}\times x\\\\OR\\\\\frac yx = \frac{m\pm tan\theta}{ 1 - \mp \,tan\theta\,}$

Hence proved

Formula used:

If θ is angle between two lines with slop m₁ and m₂ then

$tan\, \theta = \pm{ \frac { m_1-m_2}{1-m_1m_2}}$

Answer 2

$\begin{gathered}\begin{gathered}\bf \: Given - \begin{cases} &\sf{a \: line \: passes \: through \: origin \: makes} \\ &\sf{an \: angle \: \theta \: with \: y \: = mx \: + c} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: To \: prove \: \begin{cases} &\sf{equation \: of \: line \: is \: } \\ &\sf{\dfrac{y}{x} = \dfrac{m \pm tan \theta}{1 \mp m\: tan \theta} \: }  \end{cases}\end{gathered}\end{gathered}$

Concept Used

Slope of line :-

• The slope of line y = mx + c is 'm'

Angle between two lines

Let us consider two lines having slope m and M, inclined each other at angle 'p' , then angle 'p' between them is given by

$\rm :\implies\: \boxed{ \pink{ \bf \: tan \: p \: = \tt \: |\dfrac{M - M}{1 + Mm} | }}$

Slope point form of line

Let us consider a line which passes through the point (a, b) having slope 'm ', then equation of line is given by

$\rm :\implies\: \boxed{ \pink{ \bf \: y - b \: \: =m(x - a) }}$

Let's Solve the problem now!!

$\large\underline\purple{\bold{Solution :- }}$

• Let us assume that slope of the line which passes through (0, 0) is 'M'.

Now,

According to statement,

• Line makes an angle θ with line y = mx + c.

So,

$\rm :\implies\:tan \theta \: = |\dfrac{M - m}{1 + Mm} |$

$\rm :\implies\: \pm \: tan \theta \: = \dfrac{M - m}{1 + Mm}$

$\tt \: tan \theta \: = \dfrac{M - m}{1 + Mm} \: \red{or} \: \: - tan \theta \: = \dfrac{M - m}{1 + Mm}$

• Let us consider part (1)

$\rm :\implies\:tan \theta \: = \dfrac{M - m}{1 + Mm}$

$\rm :\implies\:tan \theta \: + Mmtan \theta \: = M - m$

$\rm :\implies\:tan \theta \: + m \: = M - mMtan \theta \:$

$\rm :\implies\:tan \theta \: + m \: = M(1 - mtan \theta \: )$

$\rm :\implies\: \boxed{ \pink{ \bf \: M \: = \tt \: \dfrac{tan \theta \: + m}{1 - mtan \theta \: } }} - - (1)$

Now,

• Consider part (2),

we have

$\rm :\implies\: - tan \theta \: = \dfrac{M - m}{1 + Mm}$

$\rm :\implies\: - tan \theta \: - mMtan \theta \: = M - m$

$\rm :\implies\:M + Mmtan \theta \: = m - tan \theta \:$

$\rm :\implies\:M(1 + mtan \theta \: ) = m - tan \theta \:$

$\rm :\implies\: \boxed{ \pink{ \bf \: M \: = \tt \: \dfrac{m - tan \theta \: }{1 + mtan \theta \: } }} - - (2)$

So,

• on combining (1) and (2), we get

$\rm :\implies\: \boxed{ \purple{ \bf \: M \: = \dfrac{m \pm tan \theta}{1 \mp m\: tan \theta}}}$

So,

• required equation of line which passes through (0, 0) and having slope 'M' is given by

$\rm :\implies\:y \: - \: 0 \: = \dfrac{m \pm tan \theta}{1 \mp m\: tan \theta}(x \: - \: 0)$

$\rm :\implies\:y \: = \: \dfrac{m \pm tan \theta}{1 \mp m\: tan \theta} \: x$

$\large\boxed{ \green{ \rm \: Hence, \: equation \: is \: \dfrac{y}{x} = \dfrac{m \pm tan \theta}{1 \mp m\: tan \theta}}}$

Additional Information

Additional Information Different forms of equations of a straight line

1. Equations of horizontal and vertical lines

• Equation of the lines which are horizontal or parallel to the X-axis is y = a, where a is the y – coordinate of the points on the line.
• Similarly, equation of a straight line which is vertical or parallel to Y-axis is x = a, where a is the x-coordinate of the points on the line

2. Point-slope form equation of line

• Consider a non-vertical line L whose slope is m, A(x,y) be an arbitrary point on the line and P(a, b) be the fixed point on the same line. Equation of line is given by y - b = m(x - a)

3. Slope-intercept form equation of line

• Consider a line whose slope is m which cuts the Y-axis at a distance ‘a’ from the origin. Then the distance a is called the y– intercept of the line. The point at which the line cuts y-axis will be (0,a). Then equation of line is given by y = mx + a.

4. Intercept Form of Line

• Consider a line L having x– intercept a and y– intercept b, then the line passes through  X– axis at (a,0) and Y– axis at (0,b).
• Equation of line is given by x/a + y/b = 1.

5. Normal form of Line

• Consider a perpendicular from the origin having length p to line L and it makes an angle β with the positive X-axis. Then, equation of line is given by
• x cosβ + y sinβ = p.