Question

If - 4, 3 and 4, 3 at the vertices of equilateral triangle find the coordinates of the third vertex

Answer 1

let the vertices be (x,y)
then distance between (x,y) & (4,3) is
=
\sqrt{ {(x - 4)}^{2} + {(y - 3)}^{2} }(x−4)2+(y−3)2​ 
......(1)

and distance between (x,y) & (-4,3) is
=
\sqrt{ {(x + 4)}^{2} + {(y - 3)}^{2} }(x+4)2+(y−3)2​ 
..........(2)
distance between (4,3) &(-4,3) is
=
\sqrt{ {(4 + 4)}^{2} + {(3 - 3)}^{2} }(4+4)2+(3−3)2​ 
=√(8)²=8

then
(1)=(2)
or (x-4)²=(x+4)²
or x²-8x+16=x²+8x+16
or 16x=0
or x=0

again 
(1)=8
or (x-4)²+(y-3)²=64.........(3)
putting the value of x in (3)
then (0-4)²+(y-3)²=64
or (y-3)²=64-16
or (y-3)²=48
or y-3=(+-)4√3
or y=3(+-)4√3
if we choose y as 3+4√3 then origin isn't lies interior of triangle
So required vertex is(0,3-4√3).....(ans)

Answer 2

Answer:

let the vertices be (x,y)

then distance between (x,y) & (4,3) is

=

\sqrt{ {(x - 4)}^{2} + {(y - 3)}^{2} }(x−4)2+(y−3)2​ 

......(1)

and distance between (x,y) & (-4,3) is

=

\sqrt{ {(x + 4)}^{2} + {(y - 3)}^{2} }(x+4)2+(y−3)2​ 

..........(2)

distance between (4,3) &(-4,3) is

=

\sqrt{ {(4 + 4)}^{2} + {(3 - 3)}^{2} }(4+4)2+(3−3)2​ 

=√(8)²=8

then

(1)=(2)

or (x-4)²=(x+4)²

or x²-8x+16=x²+8x+16

or 16x=0

or x=0

again 

(1)=8

or (x-4)²+(y-3)²=64.........(3)

putting the value of x in (3)

then (0-4)²+(y-3)²=64

or (y-3)²=64-16

or (y-3)²=48

or y-3=(+-)4√3

or y=3(+-)4√3

if we choose y as 3+4√3 then origin isn't lies interior of triangle