Question

If two vertices of equilateral triangle are (3,0) and (6,0), find the third vertex

Answer 1

Answer:

Step-by-step explanation:

Let ∆ABC be the equilateral triangle

Let the coordinates of vertices of ∆ABC be A (3, 0), B (6, 0) and C (x, y).

∆ABC is an equilateral triangle.

∴ AB = BC = CA

⇒ AB2 = BC2 = CA2

⇒ [(3 – 6)2 + (0 – 0)2] = [(x – 6)2 + (y – 0)2] = [(x – 3)2 + (y – 0)2]

⇒ (–3)2 = (x – 6)2 + y2 = (x – 3)2 + y2

⇒ 9 = (x – 6)2 + y2 = (x – 3)2 + y2

∴ (x – 6)2 + y2 = (x – 3)2 + y2

⇒ (x – 6)2 = (x – 3)2

⇒ x – 6 = ± (x – 3) [By taking square root on both the sides]

⇒ x – 6 = x – 3 or x – 6 = –(x – 3)

⇒ x – x = –3 + 6 or x – 6 = –x + 3

⇒ 0 = 3

(not possible)

or x + x = 3 + 6

⇒ 2x = 9

or x = 4.5

also, (x-6)^2 /+y^2=9

= (4.5-6)^2+ y^2=9

=(-1.5)^2+Y^2=9

=2.25+Y^2=9

Y^2=9-2.25

y^2=6.75

y=+-root 6.75

y=+- 2.6

Thus, coordinates of point C are (4.5, 2.6) or (4.5, – 2.6)