Question

If $a,b,c$ are sides of a triangle and
$\left|\begin{array}{ccc}{a}^2& b^2& c^2\\ (a+1{)}^2& (b+1{)}^2& (c+1{)}^2\\ (a-1{)}^2& (b-1{)}^2& (c-1{)}^2\end{array}\right|=0$, then

• A. $\Delta ABC$ is necessarily equilateral
• B. $\Delta ABC$ is necessarily right angled isosceles
• C. $\Delta ABC$ is isosceles
• D. $\Delta ABC$ is scalene triangle

Answer 1

Answer: (C)

$\Delta ABC$ is isosceles

$\left|\begin{array}{ccc}{a}^2& b^2& c^2\\ (a+1{)}^2& (b+1{)}^2& (c+1{)}^2\\ (a-1{)}^2& (b-1{)}^2& (c-1{)}^2\end{array}\right|$
Row2 elements=Row2 elements - Row1 elements
$\left|\begin{array}{ccc}{a}^2& b^2& c^2\\ 4a& 4b& 4c\\ (a-1{)}^2& (b-1{)}^2& (c-1{)}^2\end{array}\right|$
Row3 elements=Row3 elements - Row1 elements + Row2 elements/2
$\left|\begin{array}{ccc}{a}^2& b^2& c^2\\ 4a& 4b& 4c\\ 1& 1& 1\end{array}\right|$
Col1 elements=Col1 elements - Col2 elements
Col3 elements=Col3 elements - Col2 elements
$\left|\begin{array}{ccc}{a}^2-b^2& b^2& c^2-b^2\\ 4(a-b)& 4b& 4(c-b)\\ 0& 1& 0\end{array}\right|$
taking out (a-b) and (c-b) from Col1 and Col3
$(a-b)(c-b)\left|\begin{array}{ccc}a+b& b^2& b+c\\ 4& 4b& 4\\ 0& 1& 0\end{array}\right|$
which gives
$⟹(a-b)(c-b)(a-c)=0$
This happens only when any two sides become equal
So $△$ABC is isoceles