JEE Main 2026 — Mathematics Coordinate Geometry

Let the other end of the chord be $(x_1, y_1)$(x1​,y1​).

Since the chord is bisected by the $y$y-axis, the midpoint of the chord lies on the $y$y-axis.

The midpoint of the chord joining $(1, 2)$(1,2) and $(x_1, y_1)$(x1​,y1​) is $\left(\dfrac{x_1 + 1}{2}, \dfrac{y_1 + 2}{2}\right)$(2x1​+1​,2y1​+2​).

Since it lies on the $y$y-axis, its $x$x-coordinate must be zero:

$$\dfrac{x_1 + 1}{2} = 0 \Rightarrow x_1 = -1$$2x1​+1​=0⇒x1​=−1

Since the point $(x_1, y_1)$(x1​,y1​) lies on the given circle $x^2 + y^2 + x - 3y = 0$x2+y2+x−3y=0, we substitute $x_1 = -1$x1​=−1 into the equation:

$$(-1)^2 + y_1^2 + (-1) - 3y_1 = 0$$(−1)2+y12​+(−1)−3y1​=0

$$1 + y_1^2 - 1 - 3y_1 = 0$$1+y12​−1−3y1​=0

$$y_1^2 - 3y_1 = 0$$y12​−3y1​=0

$y_1(y_1 - 3) = 0 \Rightarrow y_1 = 0$y1​(y1​−3)=0⇒y1​=0 or $y_1 = 3$y1​=3

Thus, the other ends of the two chords are $R(-1, 0)$R(−1,0) and $S(-1, 3)$S(−1,3).

The midpoint of the line segment $RS$RS is given as $(\alpha, \beta)$(α,β).

$$\alpha = \dfrac{-1 + (-1)}{2} = -1$$α=2−1+(−1)​=−1

$$\beta = \dfrac{0 + 3}{2} = \dfrac{3}{2}$$β=20+3​=23​

We need to find the value of $6(\alpha + \beta)$6(α+β):

$$6(\alpha + \beta) = 6\left(-1 + \dfrac{3}{2}\right) = 6\left(\dfrac{1}{2}\right) = 3$$6(α+β)=6(−1+23​)=6(21​)=3

Answer: $3$3