Question

The equation of the right bisector plane of the segment joining $(2,3,4)$ and $(6,7,8)$ is

• A. $x+y+z+15=0$
• B. $x+y+z-15=0$
• C. $x-y+z-15=0$
• D. None of these

Answer 1

Answer: (B)

If the given points be $A(2,3,4)$ and $B(6,7,8)$, then their mid-point $N(4,5,6)$ must lie on the plane. The direction ratios of $AB$ are i.e., $1,1,1$.

$\therefore$ The required plane passes through $N(4,5,6)$ and is normal to $AB$. Thus its equation is
$1(x-4)+1(y-5)+1(z-6)=0$
$\Rightarrow x+y+z=15$