Question #ba50a

1 Answer
Dec 30, 2015

Two possibilities: C (4+5*sqrt(3)4+53, -1+5*sqrt(3)1+53), aproximately (12.66, 7.66), or D (4-5*sqrt(3)453, -1-5*sqrt(3)153), 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)d=(19)2+(4+6)2=102+102=200.

The inclination (or slope) of the line of the segment AB is k=(y1-y2)/(x1-x2)k=y1y2x1x2=>k=(4+6)/(-1-9)=-1k=4+619=1.The inclination of a perpendicular line to the previously mentioned line is p=-1/k=-1/-1 =1p=1k=11=1.

The formula of the line (which passes through the point M) containing the third vertex is y-y_M = p*(x-x_M)yyM=p(xxM) => y+1=x-4y+1=x4 => y=x-5y=x5.

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)(x+1)2+(y4)2=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)(x+1)2+(x54)2=200
Developing this last expression:
x^2+2x+1+x^2-18x+81=200x2+2x+1+x218x+81=200
2x^2-16x+82=2002x216x+82=200
x^2-8x-59=0x28x59=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)