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Difficulty: Very hardCoordinate Geometry: Lines, Slopes, and Distance

In the xyxy-plane, line l1l_1 intersects the positive xx-axis at (a,0)(a, 0) and the positive yy-axis at (0,b)(0, b), where a>b>0a > b > 0. Line l2l_2 is perpendicular to line l1l_1 and passes through the point (a,b)(a, b). If the perpendicular distance from the origin (0,0)(0, 0) to line l1l_1 is equal to the perpendicular distance from the origin to line l2l_2, what is the value of ab\frac{a}{b}?

  1. A
    5−12\frac{\sqrt{5} - 1}{2}
  2. B
    11
  3. 5+12\frac{\sqrt{5} + 1}{2}Answer
  4. D
    2\sqrt{2}
  5. E
    22

Answer

The ratio ab\frac{a}{b} is 5+12\frac{\sqrt{5} + 1}{2}.
The option stating 5+12\frac{\sqrt{5} + 1}{2} is correct because equating the distances of both lines from the origin leads to the equation a2−ab−b2=0a^2 - ab - b^2 = 0. Expressed in terms of the ratio r=abr = \frac{a}{b}, this becomes r2−r−1=0r^2 - r - 1 = 0. Because a>b>0a > b > 0, rr must be greater than 11, yielding r=5+12r = \frac{\sqrt{5} + 1}{2}.

Step-by-Step Solution

1
Find the equation and distance from origin for line l1l_1.
The equation of l1l_1 is xa+yb=1\frac{x}{a} + \frac{y}{b} = 1, or bx+ay−ab=0bx + ay - ab = 0. The perpendicular distance from (0,0)(0,0) to l1l_1 is d1=∣−ab∣a2+b2=aba2+b2d_1 = \frac{|-ab|}{\sqrt{a^2 + b^2}} = \frac{ab}{\sqrt{a^2 + b^2}}.
Standard intercept form of a line and point-to-line distance formula.
2
Find the slope and equation of line l2l_2.
The slope of l1l_1 is −ba-\frac{b}{a}, so the slope of perpendicular line l2l_2 is ab\frac{a}{b}. Since l2l_2 passes through (a,b)(a, b), its equation is y−b=ab(x−a)y - b = \frac{a}{b}(x - a), which simplifies to ax−by−(a2−b2)=0ax - by - (a^2 - b^2) = 0.
Perpendicular lines have slopes that are negative reciprocals.
3
Calculate the distance from the origin to line l2l_2 and set d1=d2d_1 = d_2.
The distance d2=∣−(a2−b2)∣a2+b2=a2−b2a2+b2d_2 = \frac{|-(a^2 - b^2)|}{\sqrt{a^2 + b^2}} = \frac{a^2 - b^2}{\sqrt{a^2 + b^2}} (since a>b>0a > b > 0). Equating d1=d2d_1 = d_2 gives aba2+b2=a2−b2a2+b2\frac{ab}{\sqrt{a^2 + b^2}} = \frac{a^2 - b^2}{\sqrt{a^2 + b^2}}, so ab=a2−b2ab = a^2 - b^2.
The distances from the origin to both lines are specified to be equal.
4
Solve for the ratio r=abr = \frac{a}{b}.
Divide a2−ab−b2=0a^2 - ab - b^2 = 0 by b2b^2 to obtain r2−r−1=0r^2 - r - 1 = 0. Applying the quadratic formula gives r=1±52r = \frac{1 \pm \sqrt{5}}{2}. Since a>b>0a > b > 0, r>1r > 1, so r=5+12r = \frac{\sqrt{5} + 1}{2}.
Determines the unique positive value greater than 1 that satisfies the algebraic relationship.

Key Concept

Perpendicular line slopes, point-to-line distance formula, and algebraic substitution for ratio determination.
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Topics

Coordinate Geometry: Lines, Slopes, and Distance

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