Question

- Find the coordinates of the point which is equidistant from the vertices of a triangle ABC,
where A(3,-1), B(-1,-6) and C(4, -1).

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Answer 1

The point which is equidistant from the points of the triangle is (2,$\frac{-8}{3}$)

1. To calculate the equidistant point we calculate the mean of the points mentioned.
2. Calculating mean of x-coordinate we get, $\frac{3 + (-1) + 4}{3}$ = 2
3. Calculating mean of y-coordinate we get,$\frac{-1 + (-6) + (-1)}{3}$ = $\frac{-8}{3}$

Answer 2

Answer:

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