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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
avatar+4330 
+1

It's an equilateral triangle.

 

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

.
 Apr 17, 2019
 #2
avatar+19789 
0

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
avatar+106519 
+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 )

 

 

 

cool cool cool

 Apr 17, 2019
 #4
avatar+106519 
+1

Thanks, tertre and EP.....!!!!

 

 

cool cool cool

 Apr 17, 2019
 #5
avatar+19789 
0

FYI....picture:

 

 Apr 17, 2019

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