Question

(a) Two vertices of an equilateral triangle are (0.3) and (4,3) Find the third vertex of the triangle​

Answer 1

Given:

• Two vertices: A (0,3) & B (4,3) of an equilateral triangle

To Find:

• The coordinate of third vertex C.

Solution:

We know that in an equilateral triangle, the distance between any two vertices would be the same. Hence using that concept, we get,

• Distance b/w AC = Distance b/w BC

This implies,

$\implies \sqrt{(x-0)^2 + (y-3)^2} = \sqrt{(x-4)^2 + (y-3)^2}\\\\\\\text{Squaring on both sides we get,}\\\\\\\implies x^2 + (y-3)^2 = (x-4)^2 + (y-3)^2\\\\\\\text{Cancelling }(y-3)^2 \text{ from both sides we get:}\\\\\\\implies x^2 = ( x^2 - 2x + 4 )\\\\\\\implies x^2 - x^2 = -2x + 4\\\\\\\implies 0 = -2x + 4\\\\\\\implies 2x = 4\\\\\\\implies \boxed{ x = \dfrac{4}{2} = 2}$

Now let's calculate the measure of each side of the equilateral triangle.

$\implies \sqrt{ ( 0 - 4 )^2 + ( 3 - 3 )^2 }\\\\\\\implies \sqrt{16} = 4\:units$

Hence the length of each side is 4 units.

Now, we know AC = 4 units. Therefore we get:

$\implies \sqrt{( x - 0)^2 + (y-3)^2} = 4\\\\\\\text{Squaring on both sides we get:}\\\\\\\implies (2)^2 + ( y^2 - 6y + 9 ) = 16\\\\\\\implies y^2 - 6y + 9 + 4 - 16 = 0\\\\\\\implies y^2 - 6y -3 = 0$

Solving the above quadratic equation we get:

$\implies y = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\\\\\\text{Substituting the required values we get,}\\\\\\\implies y = \dfrac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-3)}}{2(1)}\\\\\\\implies y = \dfrac{6 \pm \sqrt{ 36 + 12}}{2}\\\\\\\implies y = \dfrac{6 \pm \sqrt{48}}{2}\\\\\\\implies y = \dfrac{6 \pm 4\sqrt{3}}{2}\\\\\\\implies \boxed{ y = 3 + 2\sqrt{3}\:\: ;\: \: 3 - 2\sqrt{3}}$

Hence the third vertex of the triangle is:

• $[ 2, (3 + 2\sqrt{3})]$

• $[ 2, (3 - 2\sqrt{3})]$

(There are two vertex depending on which quadrant the triangle lies.)

Answer 2

Answer is in above attachment ↑

@Itzbeautyqueen23

Hope it's helpful

Thank you :)