Straight Lines
Normal/Perpendicular form of a line
Grade 11
Question:
<p>The equation of line \( AC \) in perpendicular form, given that \( A = (a\cos\alpha, a\sin\alpha) \) and the perpendicular from the origin makes an angle \( \left(\dfrac{\pi}{4} + \alpha\right) \) with the x-axis and has length \( \dfrac{a}{\sqrt{2}} \), is:</p>
<p>\( x(\cos\alpha - \sin\alpha) + y(\cos\alpha + \sin\alpha) = a \)</p>
<p>\( x(\cos\alpha + \sin\alpha) + y(\cos\alpha - \sin\alpha) = a \)</p>
<p>\( x\cos\alpha + y\sin\alpha = a \)</p>
<p>\( x(\cos\alpha - \sin\alpha) - y(\cos\alpha + \sin\alpha) = a \)</p>
Step-by-Step Solution
Key Concept: In perpendicular form, a line is written as x·cos(ω) + y·sin(ω) = p, where ω is the angle the perpendicular from origin makes with x-axis, and p is the perpendicular distance. Substitute ω = π/4 + α and p = a/√2, then verify point A satisfies the equation.
<p><strong>Step 1:</strong> Recall the perpendicular form of a line: <br>x·cos(ω) + y·sin(ω) = p<br>where ω is the angle the perpendicular from origin makes with x-axis, and p is the perpendicular distance from origin.</p><p><strong>Step 2:</strong> Substitute given values: ω = π/4 + α and p = a/√2<br>x·cos(π/4 + α) + y·sin(π/4 + α) = a/√2</p><p><strong>Step 3:</strong> Verify by substituting point A(a·cos α, a·sin α):<br>a·cos α·cos(π/4 + α) + a·sin α·sin(π/4 + α) = a/√2</p><p><strong>Step 4:</strong> Use the cosine difference formula: cos(A - B) = cos A·cos B + sin A·sin B<br>a·cos(α - (π/4 + α)) = a·cos(-π/4) = a·(1/√2) = a/√2 ✓</p><p><strong>Step 5:</strong> The equation of line AC in perpendicular form is:<br><strong>x·cos(π/4 + α) + y·sin(π/4 + α) = a/√2</strong></p><p>∴ Answer: A</p>
Correct Answer: A