Question

Question #ba50a

Answer 1

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)`