Question

$\huge\red{\fbox{\bf{Question}}}$

Show that the equation of the line passing through the origin and making an angle θ with the line y=mx+c is
$\frac{y}{x} = \frac{m± \tan θ\ \textless \ br /\ \textgreater \ \ \textless \ br /\ \textgreater \ }{1∓ \tanθ }$
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Answer 1

Appropriate Question :-

Show that the equation of the line passing through the origin and making an angle θ with the line y=mx+c is

$\sf \: \dfrac{y}{x} \: = \: \dfrac{m \pm tan\theta}{1 \mp mtan\theta }$

$\large\underline{\sf{Solution-}}$

Given that,

• A line L passing through the origin and making an angle θ with the line y=mx+c

Let assume that

• The slope of the required line L is M.

• Slope of given line y = mx + c is m

And further

• θ is the angle between the two lines having slope M and m respectively.

We know, angle between two lines is given by

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

So, it means,

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

So, Consider

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

$\rm :\longmapsto\:tan\theta +Mm tan\theta = M - m$

$\rm :\longmapsto\:tan\theta + m = - Mm tan\theta + M$

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

$\bf\implies \:M \: = \: \dfrac{tan\theta + m}{1 - mtan\theta }$

Now, Consider

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

$\rm :\longmapsto\: - tan\theta - Mm tan\theta = M - m$

$\rm :\longmapsto\:m - tan\theta = Mm tan\theta + M$

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

$\bf\implies \:M \: = \: \dfrac{m - tan\theta}{1 + mtan\theta }$

So, we concluded that

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

Now, Equation of line which passes through the point (0, 0) having slope M using slope point form is given by

$\rm :\longmapsto\:y - 0 = M(x - 0)$

$\rm :\longmapsto\:y = Mx$

$\rm \implies\:\dfrac{y}{x} = M$

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

Hence, Proved

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More to know :-

Different forms of equations of a straight line

1. Equations of horizontal and vertical lines

Equation of line parallel to x - axis passes through the point (a, b) is x = a.

Equation of line parallel to x - axis passes through the point (a, b) is x = a.

2. Point-slope form equation of line

Equation of line passing through the point (a, b) having slope m is y - b = m(x - a)

3. Slope-intercept form equation of line

Equation of line which makes an intercept of c units on y axis and having slope m is y = mx + c.

4. Intercept Form of Line

Equation of line which makes an intercept of a and b units on x - axis and y - axis respectively is x/a + y/b = 1.

5. Normal form of Line

Equation of line which is at a distance of p units from the origin and perpendicular makes an angle β with the positive X-axis is x cosβ + y sinβ = p.