if(-4,3)and(4,3)are two vertices of an equilateral triangle find the coordinates of the third vertex given that the origin - Maths - Coordinate Geometry

Question

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

Ankush Jain · Accepted Answer

Let the co-ordinate of third vertex be (x, y)

Now Using Distance formula $BC = \sqrt{[4 - (- 4)]^{2} + (3 - 3)^{2}} = \sqrt{(4 + 4)^{2} + 0} BC = \sqrt{8^{2}} = 8$ $Now, AB = \sqrt{[x - (- 4)]^{2} + (y - 3)^{2}} AB = \sqrt{(x + 4)^{2} + (y - 3)^{2}} and AC= \sqrt{(x - 4)^{2} + (y - 3)^{2}}$ Given, Î”ABC is equilateral triangle âˆ´ AB = AC = BC Now, AB = AC $\Rightarrow \sqrt{(x + 4)^{2} + (y - 3)^{2}} = \sqrt{(x - 4)^{2} + (y - 3)^{2}}$ On Squaring both sides, we get (x + 4)2 + (y â€“ 3)2 = (x â€“ 4)2 + (y â€“ 3)2 (x + 4)2 = (x â€“ 4)2 or x2 + 16 + 8x = x2 + 16 â€“ 8x â‡’ 16x = 0 x = 0 (1) AC = BC implies that $\sqrt{(x - 4)^{2} + (y - 3)^{2}} = 8 \sqrt{(0 - 4)^{2} + (y - 3)^{2}} = 8 [Using (1)]$ On squaring both sides, we get 16 + y2 + 9 â€“ 6y = 64 y2 â€“ 6y â€“ 39 = 0 $\Rightarrow y = \frac{- (- 6) \pm \sqrt{(- 6)^{2} - 4 (1) (- 39)}}{2 (1)} = \frac{6 \pm \sqrt{36 + 156}}{2} = \frac{6 \pm \sqrt{192}}{2} y = \frac{6 \pm 8 \sqrt{3}}{2} = 3 \pm 4 \sqrt{3} ∴ y = 3 + 4 \sqrt{3} and 3 – 4 \sqrt{3}$ $y \neq 3 + 4 \sqrt{3}$ , as origin lies in the interior of the triangle Third vertex $= (x, y) = (0, 3 - 4 \sqrt{3})$

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.