Question

Vanessa draws one side of equilateral ΔABC on the coordinate plane at points A(–2, 1) and B(4,1). Select all the possible coordinates of vertex C.

Answer 1

(1, $\sqrt{27}$)  and (1, - $\sqrt{27}$)

Step-by-step explanation:

Step 1 :

The co ordinates of the given equilateral triangle are A(-2,1) and B(4,1)

The distance between these 2 points is the length of the given triangle

Distance between the 2 points is  $\sqrt{(x2-x1)^{2}+ (y2-y1)^{2}}$ = sqrt (sq(4-(-2)) + sq(1-1)) = 6

Hence the given triangle has 3 equal side of length 6 unit.

Step 2:

The length of the other side should be 6. Let (x,y) be the co-ordinate of the point C

We have then x = 1 (because the perpendicular from C to AB bisects AB we have the point C to have the x co-ordinate as 1)

Also we have the distance between the point B(4,1) and C(1,y) to be 6 as this is an equilateral triangle

Hence $\sqrt{(4-1)^{2} + (1-y)^{2 }$ = 6

=> 9 + 1 + $y^{2}$ -2 y = 6

=> $y^{2}$-2 y - 26 = 0

=> y = 2± sqrt(4+104) / 2 = 1 ± sqrt(27)

Hence the possible co ordinates of C are (1, $\sqrt{27}$)  and (1, - $\sqrt{27}$)