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$and $(x_3-x_1{)}^2+(y_3-y_1{)}^2=c^2$ then 4 $\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$ is equal to

• A. $\sqrt{abc(a+b+c)}$
• B. $\sqrt{(a+b+c{)}^4}$
• C. $2\sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}$
• D. $(a+b+c)(a^2+b^2+c^2)$

Answer 1

The correct option is C $2\sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}$

where $s=\frac{a+b+c}{2}$
4 $\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|=$ 8*

Area of the triangle
$=8\sqrt{s(s-a)(s-b)(s-c)}$
$=8\sqrt{\left(\frac{a+b+c}{2}\right)\left(\frac{b+c-a}{2}\right)\left(\frac{c+a-b}{2}\right)\left(\frac{a+b-c}{2}\right)}$

$=2\sqrt{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}$