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(A) Given,the line is $x+y-\lambda=0$,which implies $x+y=\lambda$. Since $\lambda 
eq 0$,we have $\frac{x+y}{\lambda}=1$ $(i)$. The given equation of

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) Given,the line is $x+y-\lambda=0$,which implies $x+y=\lambda$. Since $\lambda 
eq 0$,we have $\frac{x+y}{\lambda}=1$ $(i)$. The given equation of the pair of lines is $x^2+y^2-2x-4y+2=0$. To find the equation of the lines joining the origin to the points of intersection $A$ and $B$,we homogenize the equation using $(i)$: $x^2+y^2-2x(1)-4y(1)+2(1)^2=0$ $x^2+y^2-2x(\frac{x+y}{\lambda})-4y(\frac{x+y}{\lambda})+2(\frac{x+y}{\lambda})^2=0$ Multiplying by $\lambda^2$: $\lambda^2(x^2+y^2)-2x\lambda(x+y)-4y\lambda(x+y)+2(x+y)^2=0$ $\lambda^2x^2+\lambda^2y^2-2x^2\lambda-2xy\lambda-4xy\lambda-4y^2\lambda+2(x^2+y^2+2xy)=0$ $(\lambda^2-2\lambda+2)x^2 + (4-6\lambda)xy + (\lambda^2-4\lambda+2)y^2 = 0$. Since $\angle AOB=90^{\circ}$,the sum of the coefficients of $x^2$ and $y^2$ must be zero: $(\lambda^2-2\lambda+2) + (\lambda^2-4\lambda+2) = 0$ $2\lambda^2-6\lambda+4 = 0$ $\lambda^2-3\lambda+2 = 0$ $(\lambda-1)(\lambda-2) = 0$. Thus,$\lambda=1$ or $\lambda=2$. Comparing with the given options,$2$ is the correct value.

Suppose $A$ and $B$ are the points at which the line $x+y-\lambda=0$ meets the pair of straight lines $x^2+y^2-2x-4y+2=0$. If $\angle AOB=90^{\circ}$,then a value of $\lambda$ is

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