Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, −3) and B is (1, 4).
A = (3, −10)
Let the endpoints of the diameter be \(A(x, y)\) and \(B(1, 4)\).
Property used: The centre of a circle is the midpoint of its diameter.
Hence, the midpoint of \(AB\) is given as \((2, -3)\).
Midpoint formula: The midpoint of the line segment joining \((x_1, y_1)\) and \((x_2, y_2)\) is
\( \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \)
Substitution: Here, \((x_1, y_1) = (x, y)\) and \((x_2, y_2) = (1, 4)\).
So, midpoint of \(AB\) = \( \left( \frac{x + 1}{2}, \frac{y + 4}{2} \right) \).
This midpoint is equal to the given centre \((2, -3)\).
\( \left( \frac{x + 1}{2}, \frac{y + 4}{2} \right) = (2, -3) \)
Equating the corresponding coordinates:
\( \frac{x + 1}{2} = 2 \Rightarrow x + 1 = 4 \Rightarrow x = 3 \)
\( \frac{y + 4}{2} = -3 \Rightarrow y + 4 = -6 \Rightarrow y = -10 \)
Conclusion: The coordinates of point \(A\) are \((3, -10)\).