Question #ba50a

1 Answer
Rui D.
Dec 30, 2015

Two possibilities: C (#4+5*sqrt(3)#, #-1+5*sqrt(3)#), aproximately (12.66, 7.66), or D (#4-5*sqrt(3)#, #-1-5*sqrt(3)#), aproximately (-4.66, -9.66)

Explanation:

Considering points A(-1,4) and B(9,-6), the middle point is M(4, -1).

The distance AB is #d=sqrt( (-1-9)^2+(4+6)^2)=sqrt(10^2+10^2)=sqrt(200)#.

The inclination (or slope) of the line of the segment AB is #k=(y1-y2)/(x1-x2)#=>#k=(4+6)/(-1-9)=-1#.The inclination of a perpendicular line to the previously mentioned line is #p=-1/k=-1/-1 =1#.

The formula of the line (which passes through the point M) containing the third vertex is #y-y_M = p*(x-x_M)# => #y+1=x-4# => #y=x-5#.

Since the triangle is equilateral the distance between the vertex and the point A (or between the vertex and point B) must be the distance d, or:
#sqrt((x+1)^2+(y-4)^2)=sqrt(200)#
Substituting y for its value given by the equation of the perpendicular line aforementioned, we have:
#sqrt((x+1)^2+(x-5-4)^2)=sqrt(200)#
Developing this last expression:
#x^2+2x+1+x^2-18x+81=200#
#2x^2-16x+82=200#
#x^2-8x-59=0#
#Delta = 64 + 236 = 300#
#x = (8+-sqrt(300))/2#
So #x1 = 4 + 5*sqrt (3)# with #y1=-1+5*sqrt(3)#
And #x2 = 4-5*sqrt (3)# with #y2=-1-5*sqrt(3)#

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