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 an equilateral triangle
• B. $\Delta$ ABC is a right angled isosceles triangle
• C. $\Delta$ ABC is an isoscles triangle
• D. none of these

Answer 1

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

$\left|\begin{array}{ccc}{a}^2& b^2& c^2\\ (a+1{)}^2& (b+1{)}^2& (c+1{)}^2\\ (a-1{)}^2& (b-1{)}^2& (c-1{)}^2\end{array}\right|=0$
$C_1\to C_1-C_2,C_2\to C_2-C_3$
$\left|\begin{array}{ccc}(a-b)(a+b)& (b-c)(b+c)& c^2\\ (a-b)(a+b+2)& (b-c)(b+c+2)& (c+1{)}^2\\ (a-b)(a+b-2)& (b-c)(b+c-2)& (c-1{)}^2\end{array}\right|=0$
$(a-b)(b-c)\left|\begin{array}{ccc}(a+b)& (b+c)& c^2\\ (a+b+2)& (b+c+2)& (c+1{)}^2\\ (a+b-2)& (b+c-2)& (c-1{)}^2\end{array}\right|=0$
$R_3\to R_3-R_1,R_2\to R_2-R_1$
$(a-b)(b-c)\left|\begin{array}{ccc}(a+b)& (b+c)& c^2\\ 2& 2& 2c+1\\ -2& -2& -2c+1\end{array}\right|=0$
$R_2\to R_2+R_3$
$(a-b)(b-c)\left|\begin{array}{ccc}(a+b)& (b+c)& c^2\\ 0& 0& 2\\ -2& -2& -2c+1\end{array}\right|=0$
$\Rightarrow -4(a-b)(b-c)(c-a)=0$
$\Rightarrow$ $a=b$ or $b=c$ or $c=a$
Hence, triangle is isosceles