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Point on Centroid Line: h is the AM of h1, h2 | JEE

JEE Maths question with a full step-by-step solution.

Question

Through the centroid of an equilateral triangle a line parallel to the base is drawn. On this line an arbitrary interior point

P

is taken. Let

h

be the distance of

P

from the base, and

h_1, h_2

its distances from the other two sides. Then

h

is the H.M. of

h_1, h_2

h

is the G.M. of

h_1, h_2

h

is the A.M. of

h_1, h_2

correct

DNone of these

Solution

Step 1: For an interior point,

[\triangle ABC]=[\triangle PBC]+[\triangle PCA]+[\triangle PAB]

. With equal sides this gives

h+h_1+h_2=\text{altitude}.

Step 2: The centroid divides the altitude

1:2

from the base, so any point on the parallel line through it has

h=\frac13(\text{altitude})

, i.e. altitude

=3h.

Step 3:

h+h_1+h_2=3h\ \Rightarrow\ h_1+h_2=2h\ \Rightarrow\ h=\frac{h_1+h_2}{2},

the A.M. of

h_1

and

h_2.

Correct answer: (3)

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