Question

find the coordinates of the point equidistant from the points a (1,2) b (3,–4) and c (5,–6). Option- (a) (-1,4) (b) (-1,2) (c) (-1,-2) (d) (-1,-1)
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Answer 1

Answer:

The given three points are A(1,2) B(3,-4) and C(5,-6).

Let P (x, y) be the point equidistant from these three points.

So, PA = PB = PC

⇒ x^2 + 1– 2x + y^2 + 4 – 4y = x^2 + 9 -6x + y^2 + 16 + 8y = x 2 + 25– 10x + y^2 + 36 + 12y

⇒ – 2x– 4y + 5 = -6x + 8y +25= – 10x + 12y+61

– 2x– 4y + 5 = -6x + 8y +25

⇒ – 2x– 4y + 5 +6x – 8y -25=0

⇒ 4x– 12y -20=0

⇒ x– 3y – 5 =0….(i)

– 2x– 4y + 5 = – 10x + 12y+61

⇒- 2x– 4y + 5 +10x – 12y-61=0

⇒8x– 16y -56=0

⇒x– 2y -7=0….(ii)

Solving (i) and (ii)

x = 11, y = 2

Thus, the required point is (11, 2)

Answer 2

Step-by-step explanation:

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