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## Answers

From the given information, we can see that the mid-point of O and P, , lies on the arbitrary line passing through O'(1, 1).

And lines OP & O'Q are perpendicular then the product of their slopes is -1.

Hence,

→ represents a Circle ( Centre=(1, 1) Radius=√2 )

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Please see the diagram.

Let the point about which the line rotates be C(1,1). When the Line L rotates, it always passes through that with the fixed point being C.

Draw one line L through C(1,1) with a slope = m.

Image of O(0,0) is found by drawing a normal to L meeting L at Q and extending to P, such that OQ = QP.

Since we are interested in locus of P. Take its coordinates as P(x₁, y₁). After finding a relation between x₁ and y₁, we will replace them by x and y.

Now Q is the midpoint of OP. Hence, Q = (x₁/2, y₁/2)

Now slope of L1 or OP = (y₁-0)/(x₁-0) = -1/m , as it is perpendicular to L.

Slope of QC = m = (1 - y₁/2) / (1 - x₁/2)

Multiply slopes, (1 - y₁/2) / (1 - x₁ /2 ) * (y₁ / x₁) = -1

y₁ - y₁²/2 = - x₁ + x₁²/2

x₁² + y₁² - 2x₁ - 2 y₁ = 0

(x₁ - 1)² + (y₁ - 1)² = 2 = (√2)²

Locus = (x - 1)² + (y - 1)² = (√2)²

circle with C(1,1) as the center and √2 as the radius.

Let the point about which the line rotates be C(1,1). When the Line L rotates, it always passes through that with the fixed point being C.

Draw one line L through C(1,1) with a slope = m.

Image of O(0,0) is found by drawing a normal to L meeting L at Q and extending to P, such that OQ = QP.

Since we are interested in locus of P. Take its coordinates as P(x₁, y₁). After finding a relation between x₁ and y₁, we will replace them by x and y.

Now Q is the midpoint of OP. Hence, Q = (x₁/2, y₁/2)

Now slope of L1 or OP = (y₁-0)/(x₁-0) = -1/m , as it is perpendicular to L.

Slope of QC = m = (1 - y₁/2) / (1 - x₁/2)

Multiply slopes, (1 - y₁/2) / (1 - x₁ /2 ) * (y₁ / x₁) = -1

y₁ - y₁²/2 = - x₁ + x₁²/2

x₁² + y₁² - 2x₁ - 2 y₁ = 0

(x₁ - 1)² + (y₁ - 1)² = 2 = (√2)²

Locus = (x - 1)² + (y - 1)² = (√2)²

circle with C(1,1) as the center and √2 as the radius.