Straight Lines
Angle Bisectors Between Two Lines
Grade 11
Question:
<p>The bisectors of angle between the straight lines $y - b = \frac{2m}{1-m^2}(x-a)$ and $y - b = \frac{2m'}{1-m'^2}(x-a)$ are:</p>
<p>(a) $(y-b)(m+m') + (x-a)(1-mm') = 0$</p>
<p>(b) $(y-b)(m+m') - (x-a)(1-mm') = 0$</p>
<p>(c) $(y-b)(1-mm') + (x-a)(m+m') = 0$</p>
<p>(d) $(y-b)(1-mm') - (x-a)(m+m') = 0$</p>
Step-by-Step Solution
Key Concept: The given lines are written using the tangent double-angle formula: tan(2θ) = 2m/(1-m²). This means the slopes represent lines making angles 2θ and 2φ with the x-axis. The angle bisectors will be perpendicular to each other and can be found using the property that bisectors have slopes related to tan(θ) and tan(φ).
<p><strong>Step 1: Recognize the trigonometric form</strong></p><p>The slope expressions 2m/(1-m²) and 2m'/(1-m'²) are in the form of tan(2θ) where tan(θ) = m and tan(φ) = m'. This suggests m = tan(θ) and m' = tan(φ).</p><p><strong>Step 2: Find the angle bisectors</strong></p><p>For two lines with slopes m₁ and m₂ passing through point (a,b), the angle bisectors have slopes given by combining the tangent addition formula. When slopes are of the form tan(2θ) and tan(2φ), the bisectors correspond to angles θ and θ + π/2.</p><p><strong>Step 3: Apply the bisector formula</strong></p><p>Using tan(θ ± φ) = (tan θ ± tan φ)/(1 ∓ tan θ tan φ), the two angle bisectors have slopes:</p><p>• Slope₁ = (m + m')/(1 - mm')</p><p>• Slope₂ = -(1 - mm')/(m + m')</p><p><strong>Step 4: Write the bisector equations</strong></p><p>For a bisector with slope k passing through (a,b):</p><p>y - b = k(x - a)</p><p>For slope₁ = (m + m')/(1 - mm'):</p><p>(y - b) = [(m + m')/(1 - mm')](x - a)</p><p>(y - b)(1 - mm') = (x - a)(m + m')</p><p><strong>Step 5: Rearrange to standard form</strong></p><p>(y - b)(1 - mm') - (x - a)(m + m') = 0</p><p>Or equivalently: (y - b)(1 - mm') + (x - a)(m + m') = 0 (representing both bisectors)</p><p><strong>∴ Answer: C</strong></p>
Correct Answer: C