Decide the third point in a equal sided triangle

I have the point (-1,-1) and (3,1) with this i could decide the equation of the line going through theese points which is y = (1/2)x - 1/2

doing -1/x i get y=-2x+2 which is a line through the center between these points. How can i decide the remaining (2) points which will have the same distances to point b and a? I'm supposed to use the **distance formula. **

Answer sheet says it's (1+√3, -2√3) and 1-√3, 2√3)

Guest Apr 17, 2019

#1**+1 **

It's an equilateral triangle.

The distance formula is \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.\)

.tertre Apr 17, 2019

#2**+2 **

Strong Work so far! I will use your results to proceed:

The didstance between the 2 given pionts is sqrt 20

set up a circle centered at -1, -1 with radius sqrt 20

(x+1)^2 + (y+1)^2 = r^2 = 20

to find the point(s) on the given perpindicular line, sub in y = -2x+2 into the circle equation

(x+1)^2 + (-2x+2 + 1)^2 = 20 expand and simplify to get

5x^2 - 10x -10 = 0 Quadratic fomula gives x = 1 +- sqrt(3) use these values to sub into the perpindicular line equation to find the correspondine 'y' values ....

ElectricPavlov Apr 17, 2019

#3**+2 **

Here's a way to do this

Find the distance^2 between both points.....we have

D^2 = (-1 - 3)^2 + (-1 - 1)^2 = 4^2 + 2^2 = 20

Now.....construct two circles with centers at the two points and a radius of D^2

So.....we have

(x + 1)^2 + (y + 1)^2 = 20

(x - 3)^2 + ( y - 1)^2 = 20

Set these equal

(x + 1)^2 + (y + 1)^2 = (x - 3)^2 + ( y - 1)^2 simplify

x^2 + 2x + 1 + y^2 + 2y + 1 = x^2 - 6x + 9 + y^2 - 2y + 1

2x + 2y + 2 = -6x - 2y + 10 divide through by 2

x + y + 1 = -3x - y + 5

2y = - 4x + 4 divide through by 2

y = -2x + 2

y = 2 - 2x (1)

Sub this into the equation for either circle for y

(x+ 1)^2 + ( 2 - 2x + 1)^2 = 20

(x+ 1)^2 + (3 - 2x)^2 = 20

x^2 + 2x + 1 + 4x^2 - 12x + 9 = 20

5x^2 - 10x + 10 = 20 divide through by 5

x^2 - 2x + 2 = 4

x^2 - 2x = 2 complete the square on x [ you can also use the quadratic formula ]

x^2 - 2x + 1 = 3

(x - 1)^2 = 3 take both roots

x - 1 = ±√3 add 1 to both sides

x = 1±√3

So

x = 1+ √3 or x = 1 - √3

And using (1)....we can find y as either

2 - 2 [ 1+ √3] = -2√3 and 2 - 2 [ 1 - √3] = 2√3

So....the two other possible points are

( 1 +√3, -2√3 ) and ( 1 - √3, 2√3 )

CPhill Apr 17, 2019