If 3 4 − 2 3 are two vertices of an equilateral triangle then the third vertex is

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).

☛ Related Questions:

The mid-points D, E, F of the sides of a triangle ABC are (3, 4), (8, 9) and (6, 7). Find the coordi . . . .
A (6, 1), B (8, 2) and C (9, 4) are three vertices of a parallelogram ABCD. If E is the midpoint of . . . .

Answer

Verified

Hint: Draw the equilateral triangle with the given vertices and let the third vertex be (x, y). As it is an equilateral triangle all sides are of equal length and the angle between all sides is \[{{60}^{\circ }}\]. We will be using distance formula which is \[d=\sqrt{\left[ {{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}} \right]}\] where \[({{x}_{1}},\,{{y}_{1}})\] and \[({{x}_{2}},\,{{y}_{2}})\] are the coordinates of any two points.Complete step-by-step answer:Before proceeding with the question, we should know about equilateral triangles. An equilateral triangle is a triangle whose all sides are equal and the angle between all sides is \[{{60}^{\circ }}\]. Let the coordinates of A be (2, 4) and B be (2, 6) and C be (x, y).

We will first calculate the distance AB between coordinates A and B, then the distance BC between B and C and finally the distance AC between A and C \[AB=\sqrt{\left[ {{(2-2)}^{2}}+{{(6-4)}^{2}} \right]}=\sqrt{{{2}^{2}}}=2\,units.......(1)\] \[BC=\sqrt{\left[ {{(2-x)}^{2}}+{{(6-y)}^{2}} \right]}=2\,units........(2)\] \[AC=\sqrt{\left[ {{(2-x)}^{2}}+{{(4-y)}^{2}} \right]}=2\,units......(3)\]We know that all the sides are equal so now \[AB=BC=AC\] .Equating equation (2) and equation (3) that is \[BC=AC\] we get, \[\,\Rightarrow \sqrt{\left[ {{(2-x)}^{2}}+{{(6-y)}^{2}} \right]}=\sqrt{\left[ {{(2-x)}^{2}}+{{(4-y)}^{2}} \right]}...........(4)\]Now squaring both sides in equation (4) we get, \[\,\Rightarrow {{(2-x)}^{2}}+{{(6-y)}^{2}}={{(2-x)}^{2}}+{{(4-y)}^{2}}.........(5)\]Cancelling similar terms on both sides from equation (5) we get, \[\,\Rightarrow {{(6-y)}^{2}}={{(4-y)}^{2}}.........(6)\]Now expanding the remaining terms in equation (6) we get, \[\,\Rightarrow 36+{{y}^{2}}-12y=16+{{y}^{2}}-8y........(7)\]Cancelling similar terms on both sides from equation (7) and solving we get, \[\begin{align} & \,\Rightarrow 12y-8y=36-16 \\ & \,\Rightarrow 4y=20 \\ & \,\Rightarrow y=5......(8) \\ \end{align}\]Now putting value of y from equation (8) in equation (2) we get, \[\,\Rightarrow \sqrt{\left[ {{(2-x)}^{2}}+{{(6-5)}^{2}} \right]}=2\,.......(9)\]Squaring both sides and solving we get, \[\begin{align} & \,\Rightarrow {{(2-x)}^{2}}+{{(6-5)}^{2}}=4\, \\ & \,\Rightarrow {{(2-x)}^{2}}+1=4 \\ & \,\Rightarrow {{(2-x)}^{2}}=3......(10) \\ \end{align}\]Taking square root of both sides of equation (10) and solving for x we get, \[\begin{align} & \,\Rightarrow (2-x)=\pm \sqrt{3} \\ & \,\Rightarrow x=2\pm \sqrt{3} \\ \end{align}\]Hence the possible coordinates of third vertex C are \[(2+\sqrt{3},\,5)\] and \[(2-\sqrt{3},\,5)\].Hence the correct answer is option (a).Note: We have to be very careful while using distance formula because in a hurry we may make simple calculation mistakes. Alternatively, we would have substituted the value of y in equation (3) then also we would have got the same answer.

Open in App

Distance between (3, 4) and (x, y)
= $\sqrt{[(x - 3)^{2} + (y - 4)^{2}]} = \sqrt{x^{2} - 6 x + 9 + y^{2} - 8 y + 16} = \sqrt{x^{2} - 6 x + y^{2} - 8 y + 25}$ ------------------------ (1)

Distance between (-2, 3) and (x, y)
= $\sqrt{(x + 2)^{2} + (y - 3)^{2}} = \sqrt{(x + 2)^{2} + (y - 3)^{2}} = \sqrt{x^{2} + 4 x + 4 + y^{2} - 6 y + 9} = \sqrt{x^{2} + 4 x + y^{2} - 6 y + 13}$ -------------------- (2)

Equating the distances we get,
$x^{2} - 6 x + y^{2} - 8 y + 25 = x^{2} + 4 x + y^{2} - 6 y + 13$ 10x + 2y - 12 = 0 5x + y - 6 = 0

y = (6 - 5x)

Substituting the value of y in equation (1) and equating it to $\sqrt{26}$ snd squaring on both sides we get

$x^{2} - 6 x + y^{2} - 8 y + 25 = 26 \Rightarrow x^{2} - 6 x + (6 - 5 x)^{2} - 8 (6 - 5 x) + 25 = 26 \Rightarrow x^{2} - 6 x + 36 + 25 x^{2} - 60 x - 48 + 40 x + 25 = 26 \Rightarrow 26 x^{2} - 26 x - 13 = 0 \Rightarrow 2 x^{2} - 2 x - 1 = 0$