First slide

Cartesian plane

Question

If the centroid and a vertex of an equilateral triangle are (2, 3) and (4, 3) respectively, then the other two vertices of the triangle are

Moderate

A

$(1, 3 + \sqrt{3})$
B

$(2, 3 + \sqrt{3})$
C

$(1, 2 + \sqrt{3})$
D

$(2, 2 + \sqrt{3})$

Solution

G being the centroid, divides AD in the ratio 2 : 1.
Since AG = 2, $∴$ GD = 1,
$∴$ Coordinates of D, using section formula, are D(1, 3).
Now AD = 1 + 2 = 3, $∴ \tan 60^{0} = \frac{3}{B D} \Rightarrow B D = \sqrt{3}$
$∴ B \equiv (1, 3 + \sqrt{3}) a n d C \equiv (1, 3 - \sqrt{3})$

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