If the two lines and , are perpendicular, then the distance of their point of intersection from the origin is
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: D
Solution
For perpendicularity, the product of slopes = -1.
Slope of line 1: -1/(a-1), slope of line 2: -2/a²
Product: (-1/(a-1))·(-2/a²) = 2/(a²(a-1)) = -1
So 2 = -a²(a-1) = -a³ + a², giving a³ - a² + 2 = 0, which factors as (a+1)(a²-2a+2) = 0. Since a²-2a+2 has no real roots, a = -1.
Substituting a = -1 into the lines:
Line 1: x + (-2)y = 1 → x - 2y = 1
Line 2: 2x + y = 1
Solving: from line 2, y = 1 - 2x. Substituting: x - 2(1-2x) = 1 → x - 2 + 4x = 1 → 5x = 3 → x = 3/5, y = 1 - 6/5 = -1/5.
Distance from origin = √((3/5)² + (-1/5)²) = √(9/25 + 1/25) = √(10/25) = √(2/5).
Slope of line 1: -1/(a-1), slope of line 2: -2/a²
Product: (-1/(a-1))·(-2/a²) = 2/(a²(a-1)) = -1
So 2 = -a²(a-1) = -a³ + a², giving a³ - a² + 2 = 0, which factors as (a+1)(a²-2a+2) = 0. Since a²-2a+2 has no real roots, a = -1.
Substituting a = -1 into the lines:
Line 1: x + (-2)y = 1 → x - 2y = 1
Line 2: 2x + y = 1
Solving: from line 2, y = 1 - 2x. Substituting: x - 2(1-2x) = 1 → x - 2 + 4x = 1 → 5x = 3 → x = 3/5, y = 1 - 6/5 = -1/5.
Distance from origin = √((3/5)² + (-1/5)²) = √(9/25 + 1/25) = √(10/25) = √(2/5).
[JEE Main 2019 · 9 Apr · Shift II]