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(D) To find the equation of the lines joining the origin to the points of intersection of the line $x + y = 1$ and the curve $x^2 + y^2 - 2y + \lambda

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(D) To find the equation of the lines joining the origin to the points of intersection of the line $x + y = 1$ and the curve $x^2 + y^2 - 2y + \lambda = 0$,we make the curve homogeneous using the line equation.Since $x + y = 1$,we can write the equation of the curve as:$x^2 + y^2 - 2y(1) + \lambda(1)^2 = 0$Substituting $1 = x + y$:$x^2 + y^2 - 2y(x + y) + \lambda(x + y)^2 = 0$Expanding the terms:$x^2 + y^2 - 2xy - 2y^2 + \lambda(x^2 + 2xy + y^2) = 0$Grouping the terms:$x^2(1 + \lambda) + xy(-2 + 2\lambda) + y^2(1 - 2 + \lambda) = 0$$x^2(1 + \lambda) + xy(2\lambda - 2) + y^2(\lambda - 1) = 0$For the lines to be perpendicular,the sum of the coefficients of $x^2$ and $y^2$ must be zero $(A + B = 0)$:$(1 + \lambda) + (\lambda - 1) = 0$$2\lambda = 0$$\lambda = 0$

The lines joining the points of intersection of the line $x + y = 1$ and the curve $x^2 + y^2 - 2y + \lambda = 0$ to the origin are perpendicular. Then the value of $\lambda$ is:

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