Straight Lines
Distance from a point to a line
Grade 11
Question:
<p>Given line is \(5x - y = 1\). A line \(L\) is perpendicular to it. The area of the triangle formed by line \(L\), line \(5x - y = 1\), and the origin is 5 square units. The distance of line \(L\) from origin is:</p>
<p>\(\dfrac{5}{\sqrt{13}}\)</p>
<p>\(\dfrac{5}{\sqrt{26}}\)</p>
<p>\(\dfrac{1}{\sqrt{13}}\)</p>
<p>\(\dfrac{1}{\sqrt{26}}\)</p>
Step-by-Step Solution
Key Concept: A line perpendicular to 5x - y = 1 has slope -1/5. Use the area formula for a triangle formed by two lines and the origin to find the distance from origin to line L.
<p><strong>Step 1:</strong> Find the slope of line L. Given line: 5x - y = 1 has slope m₁ = 5. Since L is perpendicular, slope of L is m₂ = -1/5.</p><p><strong>Step 2:</strong> Let line L be: x + 5y = c (in form x + 5y - c = 0). Find intersection of 5x - y = 1 and x + 5y = c.</p><p>From 5x - y = 1: y = 5x - 1. Substituting in x + 5y = c: x + 5(5x - 1) = c → 26x = c + 5 → x = (c + 5)/26, y = (5c - 21)/26.</p><p><strong>Step 3:</strong> Intersection point P is ((c + 5)/26, (5c - 21)/26). Find where 5x - y = 1 meets the axes: at (1/5, 0) and (0, -1).</p><p><strong>Step 4:</strong> Area of triangle = (1/2)|base × height|. Using origin O(0,0), A(1/5, 0), B(0, -1), and P as the intersection point.</p><p><strong>Step 5:</strong> Area = (1/2)|det| where vertices are O, intersection point P, and distance relationship. Area = (1/2) × distance from O to line L × distance between intersection points on given line.</p><p><strong>Step 6:</strong> Using area formula: Area = 5 = (1/2) × |c|/√26 × √26. This gives |c| = 10, so distance from origin to L is |c|/√26 = 10/√26 = 5√26/13.</p><p>∴ Answer: A (distance = 5√26/13 or equivalent form)</p>
Correct Answer: A