A triangle has a fixed base AB that is 2 cm long. The median from A to side BC is $1\frac{1}{2}$ cm long and can have any position emanating from A. The locus of the vertex C of the triangle is:
A circle with A as centre and radius 2 cm
A circle with A as centre and radius 3 cm
A circle with radius 3 cm and centre 4 cm from B along BA
An ellipse with A as a focus
Let A be at the origin (0,0) and B at (2,0). Median from A to BC has length 1.5. Let M be the midpoint of BC. AM = 1.5. The locus of M is a circle centered at A with radius 1.5. Since M is the midpoint of BC, C = 2M - B. This transformation maps the circle of M to a circle of C with radius 3, shifted by the vector -B.
Let the coordinates be A at the origin and B at 2 cm on the x-axis, making the midpoint of BC be located at the center of the locus circle since it is the fixed median from A measuring 1.5 cm. For any vertex C at (x, y), the midpoint of BC is ((x+2)/2, y/2), which must always remain 1.5 cm from A. Applying the distance formula gives sqrt(((x+2)/2)^2 + (y/2)^2) = 1.5, and squaring both sides provides (x+2)^2 + y^2 = 9. This expands to x^2 + 4x + y^2 + 5 = 0, revealing a circle with radius 3 cm and a center located at (-2, 0), which is exactly 4 cm from point B(2, 0) along the line BA.