The circumcentre of the triangle formed by $A(1, 2), B(-2, 2), C(1, 5)$ is
- $(1, 2)$
- $(-2, 2)$
- $(1, 5)$
- $\left(\dfrac{-1}{2}, \dfrac{7}{2}\right)$
Reveal answer
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Correct answer
Explanation
Circumcentre is equidistant from vertices. Distance squared from (x, y) to (1, 2), (-2, 2), (1, 5). (x-1)^2 + (y-2)^2 = (x+2)^2 + (y-2)^2 => x^2-2x+1 = x^2+4x+4 => 6x = -3 => x = -1/2. (x-1)^2 + (y-2)^2 = (x-1)^2 + (y-5)^2 => y^2-4y+4 = y^2-10y+25 => 6y = 21 => y = 7/2.
AI explanation
Let the circumcentre be (x,y), which is equidistant from all three vertices A(1,2), B(-2,2) and C(1,5). Using the distance formula for A and B gives (x-1)^2 + (y-2)^2 = (x+2)^2 + (y-2)^2, which simplifies to x = -1/2. Equating the distances for A and C gives (x-1)^2 + (y-2)^2 = (x-1)^2 + (y-5)^2, which simplifies to y = 7/2. The circumcentre is therefore at (-1/2, 7/2).