Question

Let $A(x_1,y_1),B(x_2,y_2),C(x_3,y_3)$ be three points. Area of triangle with vertices $A,B,C$ is given by

$\frac{1}{2}\left|\Delta \right|$ where,

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

If $a=BC,b=CA,c=AB$ and $2s=a+b+c$, then ${\Delta }^2$ equals

• A. $abc$
• B. $s(s-a)(s-b)(s-c)$
• C. $\frac{abc}{4}$
• D. $4s(s-a)(s-b)(s-c)$

Answer 1

Answer: (D)

$4s(s-a)(s-b)(s-c)$

Given area$=\frac{1}{2}\left|△\right|$

Heron's formula

Area$=\sqrt{S(S-a)(S-b)(S-c)}$

$\frac{1}{2}\left|△\right|=\sqrt{S(S-a)(S-b)(S-c)}$

Squaring on both sides

$=\frac{{\left|△\right|}^2}{4}=S(S-a)(S-b)(S-c)$

${\left|△\right|}^2=4S(S-a)(S-b)(S-c)$

Option D