Question

If ${(x_1-x_2)}^2+{(y_1-y_2)}^2=a^2$, ${(x_2-x_3)}^2+{(y_2-y_3)}^2=b^2$, ${(x_3-x_1)}^2+{(y_3-y_1)}^2=c^2$ and $k\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|=(a+b+c)(b+c-a)(c+a-b)\times (a+b-c)$, then the value of $k$ is

• A. $1$
• B. $2$
• C. $4$
• D. none of these

Answer 1

Answer: (C)

$4$

Given : $k{\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|}^2=(a+b+c)(b+c-a)(c+a-b)\times (a+b-c)$ ...... $\left(i\right)$

Area of any triangle $PQR$ with vertices $(x_1,y_1),(x_2,y_2)$ and $(x_3,y_3)$ is given by
$\Delta =\frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$ ......... $\left(ii\right)$

${(x_1-x_2)}^2+{(y_1-y_2)}^2=a^2$, ${(x_2-x_3)}^2+{(y_2-y_3)}^2=b^2$, ${(x_3-x_1)}^2+{(y_3-y_1)}^2=c^2$

$⟹a,b,c$ are the length of sides of a $△PQR$

Also, area of $△PQR$ with sides $a,b,c$ is

$\Delta =\sqrt{s(s-a)(s-b)(s-c)}$ ......... $[\text{Where}s=\frac{a+b+c}{2}]$

$=\sqrt{\frac{1}{16}\left(2s\right)(2s-2a)(2s-2b)(2s-2c)}$

$\Delta =\sqrt{\left[\frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{16}\right]}$ ........... $\left(iii\right)$

From $\left(ii\right)$ and $\left(iii\right)$, we get

$\frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|=\sqrt{\left[\frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{16}\right]}$

Squaring both sides, we have

$\Rightarrow 4{\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|}^2=(a+b+c)(b+c-a)(c+a-b)(a+b-c)$

Comparing above equation with $\left(i\right)$, we get $k=4$