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Miscellaneous Exercise · Q12

Q.Show that the equation of the line passing through the origin and making an angle θ\theta with the line y=mx+cy = mx + c is yx=m±tan⁡θ1∓mtan⁡θ\dfrac{y}{x} = \dfrac{m \pm \tan\theta}{1 \mp m\tan\theta}.

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A line through the origin has slope m′m', so its equation is y=m′xy = m'x. The angle between two lines with slopes mm and m′m' is given by tan⁡θ=∣m−m′1+mm′∣\tan\theta = \left|\frac{m - m'}{1 + mm'}\right|. Solving for m′m' yields m′=m±tan⁡θ1∓mtan⁡θm' = \frac{m \pm \tan\theta}{1 \mp m\tan\theta}, which gives the required form yx=m±tan⁡θ1∓mtan⁡θ\frac{y}{x} = \frac{m \pm \tan\theta}{1 \mp m\tan\theta}.

The heart of this problem lies in understanding how the angle between two lines relates to their slopes. When two non-perpendicular lines intersect, the tangent of the acute angle between them can be expressed purely in terms of their slopes. Since our desired line passes through the origin, its equation takes the simple form y=m′xy = m'x for some slope m′m', and we need to find which values of m′m' make the angle with y=mx+cy = mx + c equal to θ\theta.

The key insight is that the angle formula connects slopes to angles, letting us translate the geometric constraint (angle =θ= \theta) into an algebraic equation for the slope.

tan⁡θ=∣m1−m21+m1m2∣\tan\theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|

where θ\theta is the acute angle between lines with slopes m1m_1 and m2m_2.

Step-by-step derivation:

  1. Set up the line through the origin.

    Any line passing through the origin has the form y=m′xy = m'x, where m′m' is the slope we need to determine. Equivalently, yx=m′\frac{y}{x} = m' (for x≠0x \neq 0).

  2. Identify the slopes.

    The given line y=mx+cy = mx + c has slope mm. Our line through the origin has slope m′m'. The angle between these two lines is θ\theta.

  3. Apply the angle-between-lines formula.

    The tangent of the angle θ\theta between two lines with slopes mm and m′m' is:

tan⁡θ=∣m−m′1+mm′∣\tan\theta = \left|\frac{m - m'}{1 + mm'}\right|

  1. Remove the absolute value. Since we're looking for both possible lines making angle θ\theta with the given line (one on each side), we write:

tan⁡θ=±m−m′1+mm′\tan\theta = \pm\frac{m - m'}{1 + mm'}

This accounts for the two possible orientations.

  1. Solve for m′m'. Starting with:

tan⁡θ=±m−m′1+mm′\tan\theta = \pm\frac{m - m'}{1 + mm'}

Multiply both sides by (1+mm′)(1 + mm'):

tan⁡θ(1+mm′)=±(m−m′)\tan\theta(1 + mm') = \pm(m - m')

Expand:

tan⁡θ+mm′tan⁡θ=±m∓m′\tan\theta + m m' \tan\theta = \pm m \mp m'

Collect terms with m′m' on one side:

mm′tan⁡θ±m′=±m−tan⁡θm m' \tan\theta \pm m' = \pm m - \tan\theta

Factor out m′m':

m′(mtan⁡θ±1)=±m−tan⁡θm'(m\tan\theta \pm 1) = \pm m - \tan\theta

Therefore:

m′=±m−tan⁡θmtan⁡θ±1m' = \frac{\pm m - \tan\theta}{m\tan\theta \pm 1}

  1. Simplify the sign convention. Notice that if we take the upper signs together: m′=m−tan⁡θmtan⁡θ+1=m−tan⁡θ1+mtan⁡θm' = \frac{m - \tan\theta}{m\tan\theta + 1} = \frac{m - \tan\theta}{1 + m\tan\theta} …

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