Question

In fig. AABC is an equilateral
triangle with coordinates B and C as
B(1, 0) & C(5,0). Find the
coordinates of vertex A.
[ Ans. A(3,2√3)]​

Answer 1

Answer:

Let A (x,y)

AB=AC=BC (equaleteral triangle)

AB=AC

by distance formula

root (x-1)square+(y-0)square=root (x-5)square+(y-0)square

root se root cancel

x square+1-2x+y square=x square+25-10x+y square

x square and y square cancel

1-2x=25-10x

-2x+10x=25-1

8x=24

x=3

BA=BC

by distance formula

root (1-3)square +(0-y)square=root (1-5)square +(0-0)square

root se root cancel

(-2) square +(-y)square =(-4)square

4+y square =16

y square =16-4

y square =12

y=root 12

y= 2 root 3

(x,y)=(3,2 root 3)

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