Question

$(x_1-x_2{)}^2+(y_1-y_2{)}^2=a^2$;
$(x_2-x_3{)}^2+(y_2-y_3{)}^2=b^2$;
$(x_3-x_1{)}^2+(y_3-y_1{)}^2=c^2$;
then find $4{\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|}^2=$

• A. $(a+b+c)(b+c-a)(c+a-b)(a+b-c)$
• B. $-(a+b+c)(b+c-a)(c+a-b)(a+b-c)$
• C. $-(a+b+c)(b+c-a)(c+a-b)(a+b-c)/2$
• D. $(a+b+c)(b+c-a)(c+a-b)(a+b-c)/2$

Answer 1

The correct option is A $(a+b+c)(b+c-a)(c+a-b)(a+b-c)$

Area of the triangle PQR with vertices $(x_1,y_1),(x_2,y_2)$ and $(x_3,y_3)$ is

$\therefore \Delta =\frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$ ....(1)

Now the area of $\Delta$ $PQR$ with sides $a,b,c$ is

$\Delta =\sqrt{s(s-a)(s-b)(s-c)}$

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

$\Delta =\sqrt{\left[\frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{16}\right]}$ ....(2)

From (1) and (2), 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

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