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# With (-1,-1) and (3,1) Decide the third point in a equal sided triangle HELP!!

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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)

Apr 17, 2019

#1
+4593
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It's an equilateral triangle.

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

Apr 17, 2019
#2
+26007
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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 ....

Apr 17, 2019
#3
+111433
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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 )

Apr 17, 2019
#4
+111433
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Thanks, tertre and EP.....!!!!

Apr 17, 2019
#5
+26007
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FYI....picture:

Apr 17, 2019