The equation of the plane which bisects the l | vedclass

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Question

(A) The plane that bisects the line segment joining two points must pass through the midpoint of that segment.Let the points be $A(2, 3, 4)$ and $B(6,

Vedclass · Accepted Answer

$x + y + z - 15 = 0$

Vedclass · Answer

(A) The plane that bisects the line segment joining two points must pass through the midpoint of that segment.Let the points be $A(2, 3, 4)$ and $B(6, 7, 8)$.The midpoint $M$ of $AB$ is given by the formula $\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right)$.Substituting the coordinates,we get $M = \left( \frac{2 + 6}{2}, \frac{3 + 7}{2}, \frac{4 + 8}{2} \right) = (4, 5, 6)$.Now,we check which of the given options is satisfied by the point $(4, 5, 6)$:For option $A$: $x + y + z - 15 = 0 \implies 4 + 5 + 6 - 15 = 15 - 15 = 0$. This is satisfied.Thus,the equation of the plane is $x + y + z - 15 = 0$.

The equation of the plane which bisects the line joining $(2, 3, 4)$ and $(6, 7, 8)$ is

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