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Question
A plane meets the co-ordinates axis in A, B, C such that the centroid of the ∆ABC is the point (α, β, γ). Show that the equation of the plane is `x/alpha + y/beta + z/ϒ` = 3
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Solution
Let the equation of the plane be `x/"a" + y/"b" + z/"c"` = 1
Then the co-ordinate of A, B, C are (a, 0, 0), (0, b, 0) and (0, 0, c) respectively.
Centroid of the ∆ABC is `((x_1 + x_2 + x_3)/3, (y_1 + y_2 + y_3)/3, (z_1 + z_2 + z_3)/3)`
i.e. `("a"/3, "b"/3, "c"/3)`
But co-ordinates of the centroid of the ∆ABC are (α, β, γ) (given).
Therefore, `alpha = "a"/3`
`beta = "b"/3`
ϒ = `"c"/3`
i.e. a = 3α, b = 3β, c = 3γ
Thus, the equation of plane is `x/alpha + y/beta + z/γ` = 3
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