Question

if (4,3) and (-4,-3) are two vertices of equilateral triangle, find the coordinates of the other point given that the origin lies in the interior of the triangle.​

Answer 1

Answer:

Let the third vertex of an equilateral triangle be (x, y). Let A (- 4, 3), B(4 3) and C (x, y). We know that, in equilateral triangle the angle between two adjacent side is 60 and all three sides are equal. Read more on Sarthaks.com - https://www.sarthaks.com/126011/if-and-are-two-vertices-equilateral-triangle-then-find-the-coordinates-of-the-third-vertex.

Answer 2

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

=

[/tex]\sqrt{ {(x + 4)}^{2} + {(y - 3)}^{2} }

(x+4)

2

+(y−3)

2 [/tex]

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)

..........(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)