Suppose that two chords, drawn from the point (1, 2) on the circle x^2 + y^2 + x - 3y = 0 are bisected by the y -axis. If the other ends of these chords are R and S , and the mid point of the line segment RS is ( , ) , then 6( + ) is equal to:

Step-by-step solution

Let the other end of the chord be (x₁, y₁) . Since the chord is bisected by the y -axis, the midpoint of the chord lies on the y -axis. The midpoint of the chord joining (1, 2) and (x₁, y₁) is ( x₁ + 1 2 , y₁ + 2 2 ) . Since it lies on the y -axis, its x -coordinate must be zero: x₁ + 1 2 = 0 x₁ = -1 Since the point (x₁, y₁) lies on the given circle x^2 + y^2 + x - 3y = 0 , we substitute x₁ = -1 into the equation: (-1)^2 + y₁^2 + (-1) - 3y₁ = 0 1 + y₁^2 - 1 - 3y₁ = 0 y₁^2 - 3y₁ = 0 y₁(y₁ - 3) = 0 y₁ = 0 or y₁ = 3 T