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

Given:

$(-4, 3)$ and $(4, 3)$ are two vertices of an equilateral triangle.

To do:

We have to find the coordinates of the third vertex given that the origin lies in the interior of the triangle.

Solution:

Let $B(-4, 3)$ and $C(4, 3)$ be the two vertices of the equilateral triangle.

Let $A(x, y) be the third vertex.

All the sides in an equilateral triangle are equal.

$AB = BC = AC$

$AB = BC$

$AB^2 = BC^2$

$(\sqrt{(-4 -x)^2 + (3-y)^2})^2 = (\sqrt{(4 + 4)^2 + (3 - 3)^2})^2$

$16 + x^2 + 8x + 9 + y^2 – 6y = 64$

$x^2 + y^2 + 8x – 6y = 39$

$AB = AC$

$AB^2 = AC^2$

$(-4 – x)^2 + (3 – y)^2 = (4 – x)^2 + (3 – y)^2$

$16 + x^2 + 8x + 9 + y^2 – 18y = 16 + x^2 – 8x + 9 + y^2 – 6y$

$16x = 0$

$x = 0$

$BC = AC$

$BC^2 = AC^2$

$(4 + 4)^2 + (3 – 3)^2 = (4 – 0)^2 + (3 – y)^2$

$64 + 0 = 16 + 9 + y^2 – 6y$

$64 = 16 + (3 – y)^2$

$(3 – y)^2 = 48$

$3 – y = \pm 4\sqrt3$

$y = 3 \pm 4\sqrt3$

Therefore, the coordinates of the third vertex when origin lies in the interior of the triangle is $(0, 3 – 4\sqrt3)$.

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Updated on: 10-Oct-2022

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