Question

If area of the triangle having vertices (a, b), (b, c) and (c, a) is D, then area of the triangle having vertices (ac– b², ab – c²), (ba – c², bc – a²) and (cb – a², ca – b²) is-

• A. 2 abcD
• B. D
• C. $\frac{\Delta}{a + b + c}$
• D. (a + b + c)²D

Answer 1

Answer: (D)

(a + b + c)²D

Answer 2

We have

D = $\frac { 1 } { 2 }$ $\left| \begin{matrix} a & b & 1 \\ b & c & 1 \\ c & a & 1 \end{matrix} \right|$ = $\frac{1}{2}$ (a² + b² + c² – ab – bc – ca)

Now, area of other triangle is

D¢ =$\frac { 1 } { 2 }$ $\left| \begin{matrix} ac - b^{2} & ab - c^{2} & 1 \\ ba - c^{2} & bc - a^{2} & 1 \\ cb - a^{2} & ca - b^{2} & 1 \end{matrix} \right|$

=$\frac { 1 } { 2 }$ $\left| \begin{matrix} (c - b)(a + b + c) & (a - c)(a + b + c) & 0 \\ (a - c)(a + b + c) & (b - a)(a + b + c) & 0 \\ bc - a^{2} & ac - b^{2} & 1 \end{matrix} \right|\lbrack R_{1} \rightarrow R_{1}–R_{2}$ $$R_{2} \rightarrow R_{2} - R_{3}\rbrack$$

= $\frac{1}{2}$ (a + b + c)² $\left| \begin{matrix} c - b & a - c & 0 \\ a - c & b - a & 0 \\ bc - a^{2} & ac - b^{2} & 1 \end{matrix} \right|$

= $\frac{1}{2}$ (a + b + c)² (a² + b² + c² – ab – bc – ca)

= (a + b + c)² D.