Question

If (-4,3)and (4,3)are two vertices of an equilateral triangle, find the co-ordinates of the third vertex.

Answer 1

Let A be (-4,3) and B be (4,3)

Since slope of line AB is 0, we can say that AB is parallel to x-axis.

Also, both A and B are equidistant from y-axis and on opposite sides. Hence the side of the triangle (AB) is 8

Also, y-axis will is the perpendicual bisector from point C to line AB

Length of the perpendicular bisector =3√2⋅8=43–√=32⋅8=43

Hence y co-ordinate of point C is −(−(43–√−3)43−3)

Hence co-ordinates of point C are (0,3–43–√)

Answer 2

Solution :---

let the Third vertices be (x,y)

then Distance between (x,y) & (4,3) is :--

→ √(x-4)² + (y-3)² ---------------- Equation (1)

and Distance between (x,y) & (-4,3) is :-----

→ √(x+4)² + (y-3)² ---------------- Equation (2)

Distance between (4,3) &(-4,3) is :-------

→ √(4+4)² + (3-3)² = 8 units. ---------------- Equation (3)

Now, since, Distance Between them all is Equal , as it is Equaliteral ∆.

so, Equation (1) = Equation (2)

→ √(x-4)² + (y-3)² = √(x+4)² + (y-3)²

→ (x-4)² = (x+4)²

→ x² - 8x + 16 = x² +8x +16

→ 16x = 0

→ x = 0

And, also , Equation (1) = Equation (3)

→ √(x-4)² + (y-3)² = 8

Squaring both sides

→ (x-4)² + (y-3)² = 64

Putting value of x = 0, now,

→ (y-3)² = 64-16

→ (y-3)² = 48

Square - root both sides now,

→ (y-3) = ±4√3

→ y = ±4√3 + 3

Now, as origin lies in the interior of the triangle,

y ≠ 3+4√3 .

∴ Third vertex = (x, y) = (0, 3 - 4√3).