Question

here is a question please answer it
question no. 50

Answer 1

In equilateral triangle all 3 sides are equal...
Take the coordinates of 3rd vertex as A (x, y)
Now apply distance formula,
Now do,
AB =AC=BC
You will get your answer....

Answer 2

Two vertices of an equilateral triangle are (3, 4) and (-2, 3)
Let the third vertex of the triangle be (x, y)
Distance between (3, 4) and (-2, 3)
=√[(-2-3)2 + (3-4)2]
= (-5)2 + (-1)2
⇒ 26

Distance between (3, 4) and (x, y)
= √[(x-3)2 + (y-4)2]
= [(x-3)2 + (y-4)2]
= [(x-3)2 + (y-4)2]
= x2 - 6x + 9 + y2 - 8y + 16 
= x2 - 6x + y2 - 8y + 25 ----eq. 1

Distance between (-2, 3) and (x, y)
= √[(x+2)2 + (y-3)2]
= [(x+2)2 + (y-3)2] = 26
= x2 +  4x + 4 + y2 - 6y + 9
= x2 + 4x + y2 - 6y + 13 --- eq. 2

Equating the distances we get,
x2 - 6x + y2 - 8y + 25 = x2 + 4x + y2 - 6y + 13
10x + 2y - 12 = 0
5x + y - 6 = 0
y = (6 - 5x)

Substituting the value of y in equation (1) and equating it to 26.

x2 - 6x + y2 - 8y + 25 = 26
⇒ x2 - 6x + (6 - 5x)2 - 8(6 - 5x) + 25 = 26
⇒ x2 - 6x + 36 + 25x2 -  60x - 48 + 40x + 25 = 26
⇒ 26x2 - 26x - 13 = 0
⇒ 2x2 - 2x - 1 = 0

Solving the quadratic equation using the quadratic formula, [-b ± √(b2 - 4ac)]/2a.

x = [2 ± √(4+8)] / 4
x = [2 ± √(12)] / 4
x = [2 ± 2√(3)] / 4
x = [1 ± √(3)] / 2

y = (6 - 5x)
   = 6 - 5 [1 ± √(3)] / 2
   = [12 - 5 ± 5√(3)] / 2
   = [7 ± 5√(3)] / 2

Hence, the coordinates of the third vertex of the equilateral triangle are ([1 ± √(3)] / 2 , [7 ± 5√(3)] / 2) .


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