Question

if 2 vertex of an eqilateral tirangle are 3,0 and 6,0 find the third vertx​

Answer 1

Step-by-step explanation:

Let , the two vertex of equilateral triangle be A ( 3 , 0 ) and B ( 6 , 0 )

The distance between them is 3 units.

Use distance formula ,

d^2 =

$\sqrt{(x2 - x1) {}^{2} + (y2 - y1) {}^{2} }$

Again ,

Taking A ( 3 , 0 ) = A ( x1 , y1 )

C ( x2 , y2 )

$9 = \sqrt{(x2 - 3) {}^{2}+ (y2 - 0) {}^{2} }$

Now you can take it from here.

<(￣︶￣)↗

Answer 2

Answer:

A(3,0) and B(6,0)B(6,0), let third vertex C(a,b)C(a,b)

=>AB^{ 2 }={ BC }^{ 2 }={ CA }^{ 2 }\\ =>{ (3) }^{ 2 }+{ (0) }^{ 2 }={ (3-a) }^{ 2 }+{ b }^{ 2 }={ (6-a) }^{ 2 }+{ b }^{ 2 }\\ =>a^{ 2 }-6a+9+{ b }^{ 2 }=36+{ a }^{ 2 }-12a+{ b }^{ 2 }=9\\ =>a=\cfrac { 27 }{ 6 } \\ =>b=\left( \cfrac { 3 }{ 2 } \right) ^{ 2 }-(3)^{ 2 }\\ =>b=-\cfrac { 27 }{ 4 } \\=>AB

2

=BC

2

=CA

2

=>(3)

2

+(0)

2

=(3−a)

2

+b

2

=(6−a)

2

+b

2

=>a

2

−6a+9+b

2

=36+a

2

−12a+b

2

=9

=>a=

6

27

​

=>b=(

2

3

​

)

2

−(3)

2

=>b=−

4

27

​

So, C\left( \cfrac { 9 }{ 2 } ,-\cfrac { 27 }{ 4 } \right)C(

2

9

​

,−

4

27

​

)