Question

ABC is an equilateral triangle in which the vertices of A is(3,2), B is (5,-6) . find the vertices of C​

Answer 1

first find distance of A and B

now equal the answer of distance A and B to the distance of A and C.

hope u like this answer

Answer 2

Step-by-step explanation:

The ordinates of both B & C are 0. 

So, BC= (abscissa of B−abscissa of C)=(5−1)=4units=a

The side of equilateral ΔABC. 

$its \: height \: = \frac{ \sqrt{3} }{2} a = \frac{ \sqrt{3} }{2} \times 4 = 2 \sqrt{3} \: units \: .$

$also \: the \: mid \: point \: of \: bc \: = ( \frac{5 + 1}{2} \: \frac{0 + 0}{2} ) = (3 \: or \: \: 0)$

Now, ΔABC is equilateral and BC is on the x-axis. 

So, the abscissa of A will be the mid point of BC.

The ordinate of A will be the height of ΔABC from the mid point of BC.

∴ The coordinates of A are (3,23)