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The base of an equilateral triangle with side 2a lies along they y-axis such that the mid point of the base is at the origin. Find vertices of the triangle. - Mathematics

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प्रश्न

The base of an equilateral triangle with side 2a lies along they y-axis such that the mid point of the base is at the origin. Find vertices of the triangle.

योग
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उत्तर

Let ABC be the given equilateral triangle with side 2a.

Accordingly, AB = BC = CA = 2a

Assume that base BC lies along the y-axis such that the mid-point of BC is at the origin.

i.e., BO = OC = a, where O is the origin.

Now, it is clear that the coordinates of point C are (0, a), while the coordinates of point B are (0, –a).

It is known that the line joining a vertex of an equilateral triangle with the mid-point of its opposite side is perpendicular.

Hence, vertex A lies on the y-axis.

On applying Pythagoras theorem to ΔAOC, we obtain

(AC)2 = (OA)2 + (OC)2

⇒ (2a)2 = (OA)2 + a2

⇒ 4a2 – a2 = (OA)2

⇒ (OA)2 = 3a2

⇒ OA = `sqrt3`

∴ Coordinates of point A = `(± sqrt(3a),0)`

Thus, the vertices of the given equilateral triangle are (0, -a) and `(sqrt(3a),0)` or (0, a), (0, -a) and `(-sqrt(3a),0)`.

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अध्याय 10: Straight Lines - Exercise 10.1 [पृष्ठ २११]

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एनसीईआरटी Mathematics [English] Class 11
अध्याय 10 Straight Lines
Exercise 10.1 | Q 2 | पृष्ठ २११

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