Question 59 of 70advanced🔧 ApplyLong Answer5 marks

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\frac{y}{x} = \frac{m \pm \tan \theta}{1 \mp m \tan \theta}
Correct Answer

Let the slope of the required line passing through the origin be MM. The given line is y=mx+cy = mx + c, which has slope mm.

We know that the acute angle θ\theta between two lines with slopes m1m_1 and m2m_2 is given by:

tan⁡θ=∣m2−m11+m1m2∣\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|

Substituting m1=mm_1 = m and m2=Mm_2 = M, we get:

tan⁡θ=∣M−m1+mM∣\tan \theta = \left| \frac{M - m}{1 + mM} \right|

This gives two cases:

M−m1+mM=tan⁡θorM−m1+mM=−tan⁡θ\frac{M - m}{1 + mM} = \tan \theta \quad \text{or} \quad \frac{M - m}{1 + mM} = -\tan \theta

Solving Case 1:

M−m=tan⁡θ(1+mM)M - m = \tan \theta(1 + mM) M−m=tan⁡θ+mMtan⁡θM - m = \tan \theta + mM\tan \theta M(1−mtan⁡θ)=m+tan⁡θM(1 - m\tan \theta) = m + \tan \theta M=m+tan⁡θ1−mtan⁡θM = \frac{m + \tan \theta}{1 - m\tan \theta}

Solving Case 2:

M−m=−tan⁡θ(1+mM)M - m = -\tan \theta(1 + mM) M(1+mtan⁡θ)=m−tan⁡θM(1 + m\tan \theta) = m - \tan \theta M=m−tan⁡θ1+mtan⁡θM = \frac{m - \tan \theta}{1 + m\tan \theta}

Since the line passes through the origin, its equation is y=Mxy = Mx, or equivalently, yx=M\frac{y}{x} = M.

Therefore, the equation of the required line is:

yx=m±tan⁡θ1∓mtan⁡θ\frac{y}{x} = \frac{m \pm \tan \theta}{1 \mp m \tan \theta}

∴ Hence proved.

Exercise: Miscellaneous Exercise on Chapter 9 | Q: 12 | (Page 22)
For More Understanding

Explanation

This question tests the application of the angle formula between two lines. The key insight is that when a line makes an angle θ\theta with another line, there are two possible lines (one on each side of the given line).

The formula tan⁡θ=∣m2−m11+m1m2∣\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right| from the textbook context is used. The absolute value creates two cases, leading to two possible slopes.

Since the line passes through the origin, its equation takes the simple form y=Mxy = Mx, which gives the required result.

Solution Steps

  1. Step 1: Let MM be the slope of the required line through the origin.

  2. Step 2: Apply the angle formula:

    tan⁡θ=∣M−m1+mM∣\tan \theta = \left| \frac{M - m}{1 + mM} \right|

  3. Step 3: Remove the modulus to get two equations:

    M−m1+mM=tan⁡θorM−m1+mM=−tan⁡θ\frac{M - m}{1 + mM} = \tan \theta \quad \text{or} \quad \frac{M - m}{1 + mM} = -\tan \theta

  4. Step 4: Solve each equation for MM:

    M=m+tan⁡θ1−mtan⁡θandM=m−tan⁡θ1+mtan⁡θM = \frac{m + \tan \theta}{1 - m\tan \theta} \quad \text{and} \quad M = \frac{m - \tan \theta}{1 + m\tan \theta}

  5. Step 5: Write the equation y=Mxy = Mx and substitute the values of MM:

    yx=m±tan⁡θ1∓mtan⁡θ\frac{y}{x} = \frac{m \pm \tan \theta}{1 \mp m \tan \theta}
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