Question

A(a,b), B(c,d) and C(e,f) are the vertices of a triangle.
i) AB + BC > AC
ii) Area of the triangle = $\left(\frac{1}{2}\right)$[a(c-e)+c(f-b)+e(b-d)]
Which of the following are true?

• A. Only (i) is true
• B. Only (ii) is true
• C. Both (i) and (ii) is true
• D. Neither (i) nor (ii) is true

Answer 1

The correct option is A

Only (i) is true

Since A, B and C are the vertices of a triangle, it follows that sum of two sides is greater than the third side. So (i) is correct.
And since the area of a triangle with vertices ($x_1$, $y_1$), ($x_2$ , $y_2$ ) and ($x_3$ , $y_3$ ) is
$\frac{1}{2}$[ $x_1$($y_2$ - $y_3$) + $x_2$( $y_3$ - $y_1$) + $x_3$($y_1$ - $y_2$)]
= $\frac{1}{2}$ [a (d - f) + c (f - b) + e (b - d)] (by putting $x_1$ = a, $y_1$ = b, $x_2$ = c, $y_2$ = d, $x_3$ = e, $y_3$ = f)
Since the above area is not the same as that given in (ii), (ii) is incorrect.