Two vertices of an equilateral triangle are (0,3) and (4,3 find the third vertex of the triangle)

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Question 1If 4,3 and 4,3 are two vertices of an equilateral triangle, then find the coordinates of the third vertex, given that the origin lies in the interior of the triangle.

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Two vertices of an equilateral triangle are 0,0 and √3, √3. Find the third vertex.

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Solution:

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

Also, the origin lies in the interior of the triangle.

We have to find the coordinates of the third vertex.

Let an equilateral triangle be ABC.

The vertices of the triangle be A(-4, 3) B(4, 3) and the third vertex is (x, y).

An equilateral triangle is a triangle in which all three sides have the same length.

An equilateral triangle is also equiangular as all three internal angles are congruent to each other and are each 60°.

So, AB = BC = AC

Considering AB = BC

The distance between two points P (x₁ , y₁) and Q (x₂ , y₂) is

√[(x₂ - x₁)² + (y₂ - y₁)²]

Distance between A(-4, 3) and B(4, 3) = √[(4 - (-4))² + (3 - 3)²]

= √[(8)² + 0]

= √64

= 8

Distance between B(4, 3) and C(x, y) = √[(x - 4)² + (y - 3)²]

8 = √[(x - 4)² + (y - 3)²]

On squaring both sides,

64 = (x - 4)² + (y - 3)²

By using algebraic identity,

(a - b)² = a² - 2ab + b²

So, (x - 4)² = x² - 8x + 16

(y - 3)² = y² - 6y + 9

Now, 64 = x² - 8x + 16 + y² - 6y + 9

x² + y² - 8x - 6y = 64 - 25

x² + y² - 8x - 6y = 39 ------------------ (1)

Considering AB = AC,

Distance between A(-4, 3) and C(x, y) = √[(x - (-4))² + (y - 3)²]

= √[(x + 4)² + (y - 3)²]

Now, 8 = √[(x + 4)² + (y - 3)²]

On squaring both sides,

64 = (x + 4)² + (y - 3)²

By using algebraic identity,

(a + b)² = a² + 2ab + b²

So, (x + 4)² = x² + 8x + 16

64 = x² + 8x + 16 + y² - 6y + 9

x² + y² + 8x - 6y = 64 - 25

x² + y² + 8x - 6y = 39 ---------------- (2)

Subtracting (1) and (2),

x² + y² + 8x - 6y - (x² + y² - 8x - 6y) = 39 - 39

On simplification,

x² - x² + y² - y² + 8x - 6y + 8x + 6y = 0

8x + 8x = 0

16x = 0

x = 0

Put x = 0 in (1)

(0)² + y² - 8(0) - 6y = 39

y² - 6y - 39 = 0

Using the quadratic formula,

y = -b∙±√(b² - 4ac)/2a

Here, a = 1, b = -6 and c = -39

So, y = 6 ± √((-6)² - 4(1)(-39))/2(1)

= 6 ± √(36 + 156)/2

= (6 ± √192)/2

y = 3 ± 4√3

Now, y = 3 - 4√3

y = 3 + 4√3

Since the origin lies in the interior of the triangle, the x coordinate of the third vertex is zero.

y = 3 + 4√3 lies outside the triangle. So, it is not possible.

y = 3 - 4√3 lies inside the triangle.

Therefore, the coordinates of the third vertex = (0, 3 - 4√3)

✦ Try This: If (0, -3) and (0, 3) are the two vertices of an equilateral triangle, find the coordinates of its third vertex.

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 7

NCERT Exemplar Class 10 Maths Exercise 7.4 Problem 1

Summary:

If (– 4, 3) and (4, 3) are two vertices of an equilateral triangle, the coordinates of the third vertex, given that the origin lies in the interior of the triangle, is (0, 3 - 4√3).

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