Question

if (-4,3) and (4,3) are two vertices of an equilateral triangle,finf the coordinates of the third vertex,given that origin lies in the interior of the triangle

Answer 1

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

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

Step-by-step explanation:

Let the co-ordinate of third vertex be (x, y) Now Using Distance formula BC = [4 - (- 4)] 2 + (3 - 3) 2 = (4 + 4) 2 + 0 BC = 8 2 = 8 Now , AB = [x - (- 4)] 2 + (y - 3) 2 AB = (x + 4) 2 + (y - 3) 2 and AC = (x - 4) 2 + (y - 3) 2 Given, ΔABC is equilateral triangle

∴ AB = AC = BC

Now, AB = AC ⇒ (x + 4) 2 + (y - 3) 2 = (x - 4) 2 + (y - 3) 2

On Squaring both sides, we get

(x + 4)2 + (y – 3)2 = (x – 4)2 + (y – 3)2

(x + 4)2 = (x – 4)2

or x 2 + 16 + 8x = x 2 + 16 – 8x

⇒ 16x = 0

x = 0 ....(1)

AC = BC implies that (x - 4) 2 + (y - 3) 2 = 8 (0 - 4) 2 + (y - 3) 2 = 8 [from (1)]

On squaring both sides, we get

16 + y 2 + 9 – 6y = 64

y 2 – 6y – 39 = 0 y = -(-6) ± (- 6) 2 - 4(1)(-39) 2(1) y = 6 ± 36 + 156 2 = 6 ± 192 2 y = 6 ± 8 3 2 = 3 ± 4 3 ∴ y = 3 + 4√3 and 3 - 4√3 y ≠ 3 + 4 √3 , as origin lies in the interior of the triangle. Third vertex = (x, y) = (0, 3 - 4√3).