The plane which bisects the line joining, the points (4, –2, 3) and (2, 4, –1) at right angles also passes through the point :

Solution

Maths sir · 3 steps

A perpendicular bisector plane needs two things: the midpoint as its point and the segment as its normal.

1. Point on plane. The plane cuts the segment in half, so it passes through the midpoint ((4+2)/2, (-2+4)/2, (3-1)/2) = (3, 1, 1).
2. Normal direction. The plane meets the segment at right angles, so the segment's direction is the plane's normal. That direction is (2-4, 4+2, -1-3) = (-2, 6, -4), which is a multiple of (1, -3, 2).

   1(x-3) - 3(y-1) + 2(z-1) = 0 ⇒ x - 3y + 2z - 2 = 0

   Point	x - 3y + 2z - 2	On the plane?
   (a) (4, 0, 1)	4 - 0 + 2 - 2 = 4	No
   (b) (0, -1, 1)	0 + 3 + 2 - 2 = 3	No
   (c) (0, 1, -1)	0 - 3 - 2 - 2 = -7	No
   (d) (4, 0, -1)	4 - 0 - 2 - 2 = 0	Yes

3. So. Only (4, 0, -1) makes the left side zero, so the plane passes through it. The answer is (d).

The idea to keep. For the plane that bisects a segment at right angles, take the midpoint as the point and the joining vector as the normal. Then test each option by substituting it into the equation.

Answer from the NTA final answer key. Solution by PrepWiser.