Find the third

Geometry Level 3

If two vertices of an equilateral triangle are $(0,0),(3,\sqrt{3})$ , find the third vertex.

Note: There are two possible vertices $(x_1,y_1),(x_2,y_2)$ . Input your answer as $(y_1 \times y_2)-(x_1+x_2)$ .

The answer is -9.

1 solution

Sai Eshwar
Oct 24, 2017

Two vertices of an equilateral triangle are (0, 0) and (3, √3).

Let the third vertex of the equilaterla triangle be (x, y)

Distance between (0, 0) and (x, y) = Distance between (0, 0) and (3, √3) = Distance between (x, y) and (3, √3)

√(x^2 + y^2) = √(32 + 3) = √[(x - 3)^2 + (y - √3)^2]

x^2 + y^2 = 12 x^2 + 9 - 6x + y^2 + 3 - 2√3y = 12 24 - 6x - 2√3y = 12 - 6x - 2√3y = - 12 3x + √3y = 6 x = (6 - √3y) / 3

⇒ [(6 - √3y)/3]^2 + y^2 = 12 ⇒ (36 + 3y2 - 12√3y) / 9 + y^2 = 12 ⇒ 36 + 3y^2 - 12√3y + 9y^2 = 108 ⇒ - 12√3y + 12y^2 - 72 = 0 ⇒ -√3y + y^2 - 6 = 0 ⇒ (y - 2√3)(y + √3) = 0 ⇒ y = 2√3 or - √3

If y = 2√3, x = (6 - 6) / 3 = 0 If y = -√3, x = (6 + 3) / 3 = 3

So, the third vertex of the equilateral triangle = (0, 2√3) or (3, -√3).

stolen from toppr

Abusha Vopalat - 5 months, 2 weeks ago

Find the third

The answer is -9.

1 solution

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