Question

A, B and C are points in the xy plane such that A(1, 2); B (5, 6) and AC=3BC. Then

• A. ABC is a unique triangle
• B. There can be only two such triangles.
• C. No such triangle is possible
• D. There can be infinite number of such triangles.

Answer 1

Answer: (D)

D

Answer 2

The condition $AC = 3BC$ implies that the locus of point C is a circle of Apollonius. The equation of this circle is derived from the distance formula: $AC^2 = 9BC^2$.

Let $A = (1, 2)$ and $B = (5, 6)$. Let $C = (x, y)$. $(x-1)^2 + (y-2)^2 = 9[(x-5)^2 + (y-6)^2]$ Expanding and simplifying this equation yields the equation of a circle: $(x - \frac{11}{2})^2 + (y - \frac{13}{2})^2 = \frac{9}{2}$

A triangle ABC can be formed if A, B, and C are not collinear. The points that are collinear with A and B and satisfy $AC/BC = 3$ are the points dividing the line segment AB internally and externally in the ratio 3:1.

1. Internal Division: $C_1 = (\frac{3 \cdot 5 + 1 \cdot 1}{3+1}, \frac{3 \cdot 6 + 1 \cdot 2}{3+1}) = (4, 5)$.
2. External Division: $C_2 = (\frac{3 \cdot 5 - 1 \cdot 1}{3-1}, \frac{3 \cdot 6 - 1 \cdot 2}{3-1}) = (7, 8)$.

These two points, $C_1$ and $C_2$, lie on the line segment AB and also on the Apollonius circle. Any other point C on the circle will not be collinear with A and B, thus forming a triangle ABC. Since a circle contains infinitely many points, and only two of them cause collinearity, there are infinitely many points C that can form a triangle with A and B.