Question

There are three points (a, x), (b, y) and (c, z) such that the straight lines joining any two of them are not equally inclined to the coordinate axes where a, b, c, x, y, z $ϵ$ R.

$\left|\begin{array}{ccc}x+a& y+b& z+c\\ y+b& z+c& x+a\\ z+c& x+a& y+b\end{array}\right|$ =0 and a + c = -b, then x, $-\frac{y}{2}$, z are in

• A. A.P
• B. G.P.
• C. H.P.
• D. A.G.P

Answer 1

The correct option is A A.P

From the given conditions, $\frac{y-x}{b-a}\ne \pm 1,\frac{z-y}{c-b}\ne \pm 1,\frac{z-x}{c-a}\ne \pm 1$
$\Rightarrow x+a\ne y+b\ne z+c$ The determinant is a symmetric one. The determinant will be equal to zero
but a + b + c = 0 (given)
$\Rightarrow x+y+z=0$
$\Rightarrow x+z=2(-\frac{y}{2})\Rightarrow x,-\frac{y}{2},z$ are in A.P.