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Worked Examples · Example 2

Q.If the angle between two lines is π4\dfrac{\pi}{4} and slope of one of the lines is 12\dfrac{1}{2}, find the slope of the other line.

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The slope of the other line is found using the tangent formula for the angle between two lines. Given tan⁡θ=∣m2−m11+m1m2∣\tan\theta = \left|\frac{m_2 - m_1}{1 + m_1 m_2}\right| with θ=π4\theta = \frac{\pi}{4} and m1=12m_1 = \frac12, solving gives two possible slopes: −13-\frac13 or 33.

Why This Works

When two lines intersect, the angle between them is related to their slopes through the tangent of that angle. The formula comes from the difference of their direction angles: if a line makes an angle α\alpha with the x-axis, its slope is tan⁡α\tan\alpha. For two lines with slopes m1m_1 and m2m_2, the acute angle θ\theta between them satisfies

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

The absolute value ensures we get the acute angle (or the smaller angle between the lines). Since π4\frac{\pi}{4} is acute, we can drop the absolute value by considering both positive and negative cases — that’s why two answers appear.

Watch out

A common mistake is to forget the absolute value and solve only one equation. The angle π4\frac{\pi}{4} could come from either m2>m1m_2 > m_1 or m2<m1m_2 < m_1, so both signs must be considered.

Step-by-Step Solution

1. Set up the equation.

We know θ=π4\theta = \frac{\pi}{4}, so tan⁡π4=1\tan\frac{\pi}{4} = 1. Let m1=12m_1 = \frac12 and m2m_2 be the unknown slope. Then

1=∣m2−121+12⋅m2∣1 = \left|\frac{m_2 - \frac12}{1 + \frac12 \cdot m_2}\right|

2. Remove the absolute value by considering two cases.

Case 1: m2−121+12m2=1\frac{m_2 - \frac12}{1 + \frac12 m_2} = 1

Case 2: m2−121+12m2=−1\frac{m_2 - \frac12}{1 + \frac12 m_2} = -1

3. Solve Case 1.

m2−121+12m2=1\frac{m_2 - \frac12}{1 + \frac12 m_2} = 1

Multiply both sides by 1+12m21 + \frac12 m_2:

m2−12=1+12m2m_2 - \frac12 = 1 + \frac12 m_2

Bring terms together:

m2−12m2=1+12m_2 - \frac12 m_2 = 1 + \frac12

12m2=32\frac12 m_2 = \frac32

m2=3m_2 = 3

4. Solve Case 2.

m2−121+12m2=−1\frac{m_2 - \frac12}{1 + \frac12 m_2} = -1

Multiply:

m2−12=−1−12m2m_2 - \frac12 = -1 - \frac12 m_2

Bring terms:

m2+12m2=−1+12m_2 + \frac12 m_2 = -1 + \frac12

32m2=−12\frac32 m_2 = -\frac12

m2=−13m_2 = -\frac13

5. Verify both are valid.

Check that the denominator 1+m1m21 + m_1 m_2 is not zero:

  • For m2=3m_2 = 3: 1+12⋅3=1+1.5=2.5≠01 + \frac12 \cdot 3 = 1 + 1.5 = 2.5 \neq 0
  • For m2=−13m_2 = -\frac13: 1+12⋅(−13)=1−16=56≠01 + \frac12 \cdot (-\frac13) = 1 - \frac16 = \frac56 \neq 0

Both are valid.

Tip

Notice that the two answers are negative reciprocals? Not exactly — 33 and −13-\frac13 are negative reciprocals, which means the two lines are perpendicular to each other. That’s a coincidence here because tan⁡π4=1\tan\frac{\pi}{4}=1 leads to m1m2=−1m_1 m_2 = -1 in one case.

✓Final answer

The slope of the other line is either −13-\frac13 or 33.

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