Question

The equation of the plane which bisects the line segment joining the points $(3,2,6)$ and $(5,4,8)$ and is perpendicular to the same line segment is

• A. $x+y+z=16$
• B. $x+y+z=10$
• C. $x+y+z=12$
• D. $x+y+z=14$
• E. $x+y+z=15$

Answer 1

Answer: (D)

$x+y+z=14$

Since, plane passes through mid-point of $(3,2,6)$ and $(5,4,8)$.
Hence, $(\frac{3+5}{2},\frac{2+4}{2},\frac{6+8}{2})$ will lie on the plane.

Also, plane is perpendicular to the line segment joining $(3,2,6)$ and $(5,4,8)$.

Thus, DR's of the normal will be $5-3,4-2,8-6$ i.e., $2:2:2$ or $1:1:1$.

Hence, required equation of the plane will be
$1(x-4)+1(y-3)+1(z-7)=0$
$\Rightarrow x+y+z=14$