Straight Lines
Perpendicular lines, area of triangle and distance
Grade 11
Question:
<p>If a line \(L\) is perpendicular to the line \(5x - y = 1\), and the area of the triangle formed by the line \(L\) and the coordinate axes is 5, then the distance of line \(L\) from the line \(x + 5y = 0\) is</p>
<p>\(\dfrac{7}{\sqrt{5}}\)</p>
<p>\(\dfrac{5}{\sqrt{13}}\)</p>
<p>\(\dfrac{7}{\sqrt{13}}\)</p>
<p>\(\dfrac{5}{\sqrt{7}}\)</p>
Step-by-Step Solution
Key Concept: A line perpendicular to 5x - y = 1 has slope -1/5 (negative reciprocal of 5). Use the intercept form with the area constraint to find the line equation, then apply point-to-line distance formula.
<p><strong>Step 1:</strong> Find the slope of line L. Since L ⊥ (5x - y = 1), and the given line has slope 5, line L has slope m = -1/5.</p><p><strong>Step 2:</strong> Let line L be: x + 5y = c (rearranging from y = -x/5 + d form). The x-intercept is c and y-intercept is c/5.</p><p><strong>Step 3:</strong> Area of triangle formed by L and coordinate axes: <br>Area = (1/2)|c| · |c/5| = |c²|/10 = 5<br>Therefore: |c²| = 50, so c = ±5√2</p><p><strong>Step 4:</strong> The two possible lines are x + 5y = 5√2 and x + 5y = -5√2.</p><p><strong>Step 5:</strong> Distance from line x + 5y = c to line x + 5y = 0 is:<br>d = |c|/√(1² + 5²) = |c|/√26 = 5√2/√26 = 5√2·√26/26 = 5√52/26 = 10√13/26 = <strong>5√13/13</strong></p><p>∴ Answer: C</p>
Correct Answer: C