Question

if (-4,0) and (4,0) are two vertices of an equilateral triangle , find the coordinates of its third vertices

Answer 1

Let A(x,y), B(-4,0), C(4,0)
Distance between BC = 8 (Calculate it)
Distance between AB = √ (-4-x)^2 + (0-y)^2
Distance between AC = √ (4-x)^2 + (0-y)^2
Given that the triangle is equilateral. So, AB=BC= AC

AB=AC
√(-4-x)^2+ (0-y)^2 = √ (4-x)^2+ (0-y)^2

(-4-x)^2+ (0-y)^2 = (4-x)^2+ (0-y)^2
(-4-x)^2= (4-x)^2
x^2+8x+16= x^2-8x+16
8x+8x= 16-16
x= 0 ..........(1)

Again, AC = BC

√(4-x)^2+ (0-y)^2 = 8
(4-x)^2+ (0-y)^2= 64
(4-0)^2+ (0-y)^2 { substituting 1}= 64
4^2 + y^2 = 64
y^2 = 64-16=38
y = +√38 or -√38

Therefore, x=0 and y = +√38 or -√38
Third vertex is (0, √38) or (0,-√38)

Answer 2

Answer:

Let C(x,y) be the third vertex of triangle ABC having two vertices at A(-4,0) and B(4,0).Since ABC is an equilateral triangle. Therefore,

AC=BC=AB

now, AC=BC

√(x+4)^2 + (y+0)^2=√(x-4) ^2 + (y-0) ^2

(x+4)^2 + y^2=(x-4) ^2 + y^2

16x=0

x=0

Again, AC=AB

√(x+4)^2+(y-0) ^2=√(4+4)^2 + 0

(x+4)^2+ y^2 = 64

(0+4)^2 + y^2= 64

y^2= 48

y= +/- 4√3

Hence, the coordinates of the third vertex are C(0,4√3)/(0,-4√3)

Step-by-step explanation: