Question

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. A straight line AB, $1\frac{1}{2}$ cm from A
• B. A circle with A as centre and radius 2 cm
• C. A circle with A as centre and radius 3 cm
• D. A circle with radius 3 cm and centre 4 cm from B along BA
• E. An ellipse with A as a focus

Answer 1

Answer: (D)

A circle with radius 3 cm and centre 4 cm from B along BA

perhaps the easiest approach to this problem is through coordinate geometry. Let the triangle be A(0, 0) , B(2, 0) and C(x, y). Then $A_1$ the midpoint of BC, has the coordinate [(x+2), y/2]. Therefore using the distance formula, we have
$\sqrt{(\frac{x+2}{2}{)}^2+(\frac{y}{2}{)}^2}=\frac{3}{2}$ or $(\frac{x+2}{2}{)}^2+(\frac{y}{2}{)}^2=\frac{9}{4}$ or $(x+2{)}^2+y^2=9$
This equation represents a circle with radius 3 cm and centre 4 cm from B along BA; or as extreme values we may have a-b = 2 and a-b = 0, using the notation a, b, c with its usual meaning. From the median formula , with c = 2 and m= 3/2 , we have $2b^2+8=9+a^2$ $\therefore$ 2b^2- a^2= 1$. Using the given extreme values in succession, we have b = 5 and a = 7, and b = 1 and a = 1. Therefore, the maximum distance between the extreme positions of C is 6 cm. These facts, properly collated, lead to the same result as that shown above.