Straight Lines
Distance from Point to Line
Grade 11
Question:
<p>A straight line L at a distance of 4 units from the origin makes positive intercepts on the coordinate axes and the perpendicular from the origin to this line makes an angle of 60° with the line <i>x + y = 0</i>. Then, an equation of the line L is</p>
<p>(a) <i>x + √3y = 8</i></p>
<p>(b) <i>(√3 + 1)x + (√3 - 1)y = 8√2</i></p>
<p>(c) <i>√3x + y = 8</i></p>
<p>(d) <i>(√3 - 1)x + (√3 + 1)y = 8√2</i></p>
Step-by-Step Solution
Key Concept: Use the normal form of a line equation. The perpendicular distance formula and angle conditions determine the inclination of the normal to the line.
<p><strong>Step 1:</strong> Let <i>θ</i> be the inclination of the line <i>x + y = 0</i>. Then <i>tan θ = -1 = tan(180° - 45°)</i>, so <i>θ = 135°</i>.</p><p><strong>Step 2:</strong> Since the perpendicular from the origin to line L makes an angle of 60° with the line <i>x + y = 0</i>, we have <i>α + 60° = 135°</i>, where <i>α</i> is the inclination of the perpendicular.</p><p><strong>Step 3:</strong> This gives <i>α = 75°</i>.</p><p><strong>Step 4:</strong> The equation of line L with perpendicular distance OM = 4 is <i>x cos 75° + y sin 75° = 4</i>.</p><p><strong>Step 5:</strong> Simplifying with <i>cos 75° = (√3 - 1)/(2√2)</i> and <i>sin 75° = (√3 + 1)/(2√2)</i>, we get <i>(√3 - 1)x + (√3 + 1)y = 8√2</i>.</p><p>∴ Answer is (d).</p>
Correct Answer: D