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. $\Delta ABC$ s an equilateral triangle
• B. $\Delta ABC$ is right angled isosceles triangle
• C. $\Delta ABC$ is an isosceles triangle
• D. $\Delta ABC$ None of the above

Answer 1

The correct option is C $\Delta ABC$ is an isosceles triangle

$\Delta =\left|\begin{array}{ccc}{a}^2& b^2& c^2\\ (a+1{)}^2& (b+1{)}^2& (c+1{)}^2\\ (a-1{)}^2& (b-1{)}^2& (c-1{)}^2\end{array}\right|$
Applying $R_2\to R_2-R_3$
$=4\left|\begin{array}{ccc}{a}^2& b^2& c^2\\ a& b& c\\ (a-1{)}^2& (b-1{)}^2& (c-1{)}^2\end{array}\right|$
Applying $R_3\to R_3-R_1+2R_2$
$\therefore \Delta =4\left|\begin{array}{ccc}{a}^2& b^2& c^2\\ a& b& c\\ 1& 1& 1\end{array}\right|=-4(a-b)(b-c)(c-a)=0$
If a – b = 0 or b – c = 0 or c – a = 0
$\therefore$ a = b or b = c or c = a
$\therefore \Delta ABC$ is an isosceles triangle.