Question

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

Answer 1

(0, 4)....
hope it will help you....

Answer 2

Answer:

The third vertex is $(0,\pm4\sqrt3)$

Step-by-step explanation:

Given if B(-4,0) and C(4,0) are two vertices of an equilateral triangle. we have to find the coordinates of its third vertex.

Let A(x,y) be the coordinate of its third vertex.

Using distance formula,

$\text{if }(x_1,y_1) \text{and }(x_2,y_2) \text{are the two coordinate points then distance between these two }$

$Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$BC =\sqrt{(4-(4))^2+(0-0)^2}=\sqrt{64+0}= 8$

$AB = \sqrt{(-4-x)^2 + (0-y)^2}$

$AC =\sqrt{(4-x)^2 + (0-y)^2}$

Given that the triangle is equilateral. So, AB=BC= AC

∴ AB=AC

$\sqrt{(-4-x)^2+ (0-y)^2} =\sqrt{ (4-x)^2+ (0-y)^2}$

⇒ $(-4-x)^2+ (0-y)^2 = (4-x)^2+ (0-y)^2$

⇒ $(-4-x)^2= (4-x)^2$

⇒ $x^2+8x+16= x^2-8x+16$

⇒ $16x=0$ ⇒ x=0

Again, AC = BC

$\sqrt{(4-x)^2+ (0-y)^2} = 8$

Squaring on both sides, we get

$(4-x)^2+ (0-y)^2= 64$

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

$4^2 + y^2 = 64$

$y^2 = 64-16=48$

$y = \pm4\sqrt3$

Hence, the third vertex is $(0,\pm4\sqrt3)$