A plane meets the co-ordinate axes in A,B,C such that the centroid of the triangle ABC is the point $(p, q, r)$ . The equation of the plane is

\frac{x}{p} + \frac{y}{q} + \frac{z}{r} = 0

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\frac{x}{p} + \frac{y}{q} + \frac{z}{r} = 1

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\frac{x}{p} + \frac{y}{q} + \frac{z}{r} = 2

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none of these

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Solution

The correct option is A none of these
The equation of plane is

$\frac{x}{A} + \frac{y}{B} + \frac{z}{C} = 1$ ……. (1)

Using centroid formula in XYZ plane is

$(x = \frac{x_{1} + x_{2} + x_{3}}{3}, y = \frac{y_{1} + y_{2} + y_{3}}{3}, z = \frac{z_{1} + z_{2} + z_{3}}{3})$

Then,

$(\frac{A}{3}, \frac{B}{3}, \frac{C}{3}) = (p, q, r)$

Therefore,

$A = 3 p, B = 3 q, C = 3 r$

Put the value of $A, B$ and $C$ in equation of plane (1) and we get

$\frac{x}{A} + \frac{y}{B} + \frac{z}{C} = 1$

$\frac{x}{3 p} + \frac{y}{3 q} + \frac{z}{3 r} = 1$

Hence,

$\frac{x}{p} + \frac{y}{q} + \frac{z}{r} = 3$
option (D) is correct answer.

Suggest Corrections

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