Question

Two vertices of an equilatral triangle are (0,3) and (4,3). Find the third vertex of the triangle .

Answer 1

By using distance formula side AB=
(0-4)^2 +(3-3)^2 whole root
16roit
=4
Now all side of equilateral triangle is equal
So let sideAB=BC=AC and also let side AC(X, Y)

Ac=(0-X)^2 +(3-y)^2 whole under root
(4)^2 =X^2 +Y^2 +9-6Y
X^2 +Y^2 - 6y+9-16=0
X^2+ Y^2 - 6Y-7=0. . . . . (1)

Again side BC Distance
(4) ^2=(4-x)^2 +(3-y)^2
16=16+X^2 - 8X+9+y^2 - 6y
X^2 - 8X+y^2 +9-6y=0. . . .( 2)

Now from 1, 2 equation
X^2 +y^2 - 6y-7=X^2 - 8X+Y^2 +9-6y
X=2. . . By calculate

Then on (1), (2) Y also be calculate
Y=-3+-2root3
Hence c(2,-3+-2root3
I hope this answer help you,
For any mistakes sorry.
On (1)substitute by using quadric equation
See, (2)^2 +Y^2 - 6y-7
y^2 - 6Y-3
-6+-whole under root 36+12/2*1
=-3+-2under root 3

Answer 2

Answer:

Step-by-step explanation:

see the pic