Question

Let BC be fixed line segment in theplane . The locus of a point A such that the triangle ABC is isosceles is
a ) a line
b ) a circle
c ) the union of a circle and a line
d ) the union of two circles and a line

Answer 1

Here the locus is a line which passes through the mid point of the line BC. Hence Option A would be correct

Answer 2

Thank you for asking this question. Here is your answer:

The correct answer for this question is OPTION D

If ∠B = ∠C

locus of A is⊥ bisector of BC

So it is straight line

If ∠A = ∠C

BC fixed B(a, 0), C(0, a)

BC = AB

So, (x – a)2

+ y2

= 2a2

Circle

∠A = ∠B

AC = BC  

√h² + (k- a²) =  √2 a ²

x ²  + (y – a)² = 2a²

Also a circle

So union of two circle and a line.

If there is any confusion please leave a comment below.