Question

If the point (x, y) on the tangent is equidistant from the points (2, 3) and (6, - 1), find the relation between x and y.

Answer 1

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Answer :

(3, 7), (6, 5) and (15, –1)

=>Let the points be A (15, –1), B (6, 5) and C (3, 7)

Distance of AB

⇒ AB = √ (6 – 15)2 + (5 – (–1))2

⇒ AB = √ (–9)2 + (6)2

⇒ AB = √ (81 + 36)

⇒ AB = √ 117 = √ 3 × 3 × 13

⇒ AB = 3√13

Distance of BC

⇒ BC = √ (3 – 6)2 + (7 – 5)2

⇒ BC= √ (3)2 + (2)2

⇒ BC = √ (9 + 4)

⇒ BC= √ 13

Distance of AC

⇒ AC = √ (3 – 15)2 + (7 – (–1))2

⇒ AC = √ (3 – 15)2 + (7 + 1)2

⇒ AC= √ (–12)2 + (8)2

⇒ AC = √ (144 + 64)

⇒ AC= √ 208 = √ 4 × 4 × 13

⇒ AC = 4√13

i.e. AB + BC = AC

⇒ 3√13 + √13 = 4√13

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Answer 2

Let the points be P(x,y), A(6,-1) and B(2,3).

AP2 = (x-6)2 + (y+1)2

BP2 = (x-2)2 + (y-3)2

Given, 9(x,y) is equidistant from (6,-1) and (2,3)

(x-6)2 + (y+1)2 = (x-2)2 + (y-3)2

x2 - 12x + 36 + y2 + 2y + 1 = x2 - 4x + 4 + y2 - 6y + 9

-12x + 36 + 2y + 1 = - 4x + 4 - 6y + 9

-8x + 8y = -24 

-x + y = -3

x - y = 3. 

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