Search any question & find its solution
Question:
Answered & Verified by Expert
Let $\mathrm{BC}$ be a fixed line segment in the plane. The locus of a point $A$ such that the triangle $\mathrm{ABC}$ is isosceles, is (with finitely many possible exceptional points)
Options:
Solution:
1195 Upvotes
Verified Answer
The correct answer is:
the union of two circles and a line
Case (i):
If $\angle \mathrm{B}=\angle \mathrm{C}$
locus of $\mathrm{A}$ is $\perp$ bisector of $\mathrm{BC}$
So it is straight line
Case (ii):
If $\angle \mathrm{A}=\angle \mathrm{C}$
BC fixed $\mathrm{B}(\mathrm{a}, 0), \mathrm{C}(0, \mathrm{a})$
$\mathrm{BC}=\mathrm{AB}$
So, $(x-a)^{2}+y^{2}=2 a^{2}$
Circle
Case (iii):
$\begin{array}{l}
\angle \mathrm{A}=\angle \mathrm{B} \\
\mathrm{AC}=\mathrm{BC} \\
\sqrt{\mathrm{h}^{2}+(\mathrm{k}-\mathrm{a})^{2}}=\sqrt{2 \mathrm{a}^{2}} \\
\mathrm{x}^{2}+(\mathrm{y}-\mathrm{a})^{2}=2 \mathrm{a}^{2}
\end{array}$
also a circle
So union of two circle and a line.
If $\angle \mathrm{B}=\angle \mathrm{C}$
locus of $\mathrm{A}$ is $\perp$ bisector of $\mathrm{BC}$
So it is straight line
Case (ii):
If $\angle \mathrm{A}=\angle \mathrm{C}$
BC fixed $\mathrm{B}(\mathrm{a}, 0), \mathrm{C}(0, \mathrm{a})$
$\mathrm{BC}=\mathrm{AB}$
So, $(x-a)^{2}+y^{2}=2 a^{2}$
Circle
Case (iii):
$\begin{array}{l}
\angle \mathrm{A}=\angle \mathrm{B} \\
\mathrm{AC}=\mathrm{BC} \\
\sqrt{\mathrm{h}^{2}+(\mathrm{k}-\mathrm{a})^{2}}=\sqrt{2 \mathrm{a}^{2}} \\
\mathrm{x}^{2}+(\mathrm{y}-\mathrm{a})^{2}=2 \mathrm{a}^{2}
\end{array}$
also a circle
So union of two circle and a line.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.