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. $x +y+z =14$

Answer 1

Answer: (D)

$x +y+z =14$

Answer 2

Since, plane passes through mid-point of $(3,2,6)$ and $(5,4,8)$. Hence, $\left(\frac{3+5}{2}, \frac{2+4}{2}, \frac{6+8}{2}\right)$ 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 $1(x-4)+1(y-3)+1(z-7)=0$ $\Rightarrow x+y+z=14$