Question

In the given fig. ∆ABC is an equilateral triangle of side 3 units. Find the coordinates of the other 2 vertices.

Answer 1

Answer:

Side = 3 units

Coordinate B:

y axis = 0

x axis = 2 + 3 = 5

B = (5,0)

Coordinate C:

Midpoint formula: (x,y) = ((x1 + x2)/2 , (y1 + y2)/2)

Midpoint of AB: (x,y) = ((2 + 5)/2 , (0 + 0)/2)

=> (x,y) = (7/2 , 0) = (3.5,0)

Thus, the coordinate C lies on the point (3.5 , y)

When distance of this point from A is 3 units:

Distance formula: $\tt{\sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}}}$

=> $\tt{\sqrt{(3.5 - 2)^{2} + (y - 0)^{2}}}$ = 3

=> $\tt{\sqrt{2.25 + y^{2}}}$ = 3

Squaring on both sides:

=> 2.25 + y^2 = 9

=> y^{2} = 9 - 2.25

=> y^2 = 6.75

=> y = 2.5

Coordinate C = (3.5, 2.5)

Coordinate B = (5,0)

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