Question

If ${△}_1,{△}_2$ be the areas of two triangles with vertices $(b,c),(c,a),(a,b)$, and $(ac-b^2,ab-c^2),(ba-c^2,bc-a^2),(cb-a^2,ca-b^2)$, then $\frac{{△}_1}{{△}_2}=(a+b+c{)}^2$

• A. True
• B. False

Answer 1

The correct option is A True

$T_1=(b,c)(c,a)(a,b)=(x_1,y_1)(x_2,y_2)(x_3,y_3)$

$T_2=\left[\right(ac-b^2),(ab-c^2\left)\right],\left[\right(ba-c^2),(bc-a^2\left)\right],\left[\right(cb-a^2),(ca-b^2\left)\right]$

$T_1=\frac{1}{2}\left[c\right(c-a)+a(a-b)+b(b-c\left)\right]$

$=\frac{1}{2}[c^2-ac+a^2-ab+b^2-bc]$

$T_2=\frac{1}{2}\left[\right(ab-c^2\left)\right[ba-c^2-cb+a^2]+(bc-a^2\left)\right[cb-a^2-ac+b^2]+(ca-b^2\left)\right[ac-b^2-ba+c^2\left)\right]$

$=\frac{1}{2}\left[\right[(ab-c^2)(a-c)(a+b+c)]+(bc-a^2\left)\right[(b-a)(a+b+c)]+(ca-b^2\left)\right[(c-b)(a+b+c)\left]\right]$

$=\frac{(a+b+c)}{2}[a^{2}b-abc-a{c}^2+c^3+b^{2}c-abc-a^{2}b+c^{2}a-abc-b^{2}c+b^3]$

$=\frac{(a+b+c)}{2}(a^3+b^3+c^3-3abc)$

$=\left(\frac{a+b+c}{2}\right)[a+b+c][a^2+b^2+c^2-ab-bc-ca]$

$\therefore \frac{{\Delta }_1}{{\Delta }_2}=\frac{(a+b+c{)}^2[a^2+b^2+c^2-bc-ba-ca]}{(a^2+b^2+c^2-bc-ba-ca)}$

$=(a+b+c{)}^2$