Question:
Consider a triangle having vertices A\((-2, 3)\), B\((1, 9)\) and C\((3, 8)\). If a line L passing through the circum-centre of triangle ABC, bisects line BC, and intersects y-axis at point
\(\left(0, \frac{\alpha}{2}\right)\)
, then the value of real number \(\alpha\) is ______________.
Consider a triangle having vertices A\((-2, 3)\), B\((1, 9)\) and C\((3, 8)\). If a line L passing through the circum-centre of triangle ABC, bisects line BC, and intersects y-axis at point
\(\left(0, \frac{\alpha}{2}\right)\)
, then the value of real number \(\alpha\) is ______________.
\(\left(0, \frac{\alpha}{2}\right)\)
, then the value of real number \(\alpha\) is ______________.
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Find the circumcentre O by equating distances, take the midpoint M of BC, and write the line OM.
Updated On: Nov 4, 2026
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Correct Answer: 9
Solution and Explanation
Let circumcentre O = (x, y). \(OA^2 = OB^2\): \((x+2)^2 + (y-3)^2 = (x-1)^2 + (y-9)^2 \Rightarrow 6x + 12y = 69 \Rightarrow 2x + 4y = 23\).
\(OB^2 = OC^2\): \((x-1)^2 + (y-9)^2 = (x-3)^2 + (y-8)^2 \Rightarrow 4x - 2y = -9 \Rightarrow y = 2x + \frac{9}{2}\).
Substituting: \(2x + 8x + 18 = 23 \Rightarrow x = \frac{1}{2}\), \(y = \frac{11}{2}\). O = (1/2, 11/2).
Midpoint of BC: M = (2, 17/2). Slope of OM = \(\frac{17/2 - 11/2}{2 - 1/2} = \frac{3}{3/2} = 2\).
Line L: \(y - \frac{11}{2} = 2\left(x - \frac{1}{2}\right) \Rightarrow y = 2x + \frac{9}{2}\). At x = 0, \(y = \frac{9}{2} = \frac{\alpha}{2}\), so \(\alpha = 9\).
Answer: 9
\(OB^2 = OC^2\): \((x-1)^2 + (y-9)^2 = (x-3)^2 + (y-8)^2 \Rightarrow 4x - 2y = -9 \Rightarrow y = 2x + \frac{9}{2}\).
Substituting: \(2x + 8x + 18 = 23 \Rightarrow x = \frac{1}{2}\), \(y = \frac{11}{2}\). O = (1/2, 11/2).
Midpoint of BC: M = (2, 17/2). Slope of OM = \(\frac{17/2 - 11/2}{2 - 1/2} = \frac{3}{3/2} = 2\).
Line L: \(y - \frac{11}{2} = 2\left(x - \frac{1}{2}\right) \Rightarrow y = 2x + \frac{9}{2}\). At x = 0, \(y = \frac{9}{2} = \frac{\alpha}{2}\), so \(\alpha = 9\).
Answer: 9
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