Q.21. Identify the curve

Question 21.
Identify the curve √[x^2+(y-1)^2] + √[(x-1)^2+y^2] = √2.

Answer 21.
√[x^2+(y-1)^2] + √[(x-1)^2+y^2] = √2 ... (1)

Rationalising LHS,
[x^2 + (y-1)^2 - (x-1)^2 - y^2] / [√[x^2+(y-1)^2] - √[(x-1)^2+y^2]] = √2
=> 2(x - y) / [√[x^2+(y-1)^2] - √[(x-1)^2+y^2]] = √2
=> √[x^2+(y-1)^2] - √[(x-1)^2+y^2] = √2 (x - y) ... (2)

Adding (1) and (2),
2√[x^2+(y-1)^2] = √2 + √2(x - y)
=> √2 * √[x^2+(y-1)^2] = (1 + x - y)
=> 2(x^2 + y^2 - 2y + 1) = 1 + x^2 + y^2 - 2xy + 2x - 2y
=> x^2 + y^2 + 2xy - 2x - 2y + 1 = 0
=> (x + y - 1)^2 = 0

Whenever we square both sides of an equation, we introduce one different equation also as a question.
The question is
√[x^2+(y-1)^2] + √[(x-1)^2+y^2] = √2
Since none of the two terms on LHS can be negative, each has to be less than or equal to √2
=> x^2 + (y-1)^2 ≤ 2
and (x-1)^2 + y^2 ≤ 2
=> all points on the graph of the given equation shlould lie within the common area of the above two circles
=> (1, 0) and (0, 1) are the end points of the concident line
x + y = 1
=> the given equation represents a line-segment.

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