(A) Given the line $x+2y=k$,we have $\frac{x+2y}{k}=1$. Substituting this into the equation of the curve $2x^2-2xy+3y^2+(2x-y)(1)-(1)^2=0$,we get: $2x

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(A) Given the line $x+2y=k$,we have $\frac{x+2y}{k}=1$. Substituting this into the equation of the curve $2x^2-2xy+3y^2+(2x-y)(1)-(1)^2=0$,we get: $2x^2-2xy+3y^2+(2x-y)\left(\frac{x+2y}{k}\right)-\left(\frac{x+2y}{k}\right)^2=0$. Multiplying by $k^2$,we obtain: $k^2(2x^2-2xy+3y^2)+k(2x^2+4xy-xy-2y^2)-(x^2+4xy+4y^2)=0$. Expanding and grouping the terms: $x^2(2k^2+2k-1) - xy(2k^2-3k+4) + y^2(3k^2-2k-4) = 0$. Since $OA \perp OB$,the sum of the coefficients of $x^2$ and $y^2$ must be zero: $(2k^2+2k-1) + (3k^2-2k-4) = 0$. $5k^2 - 5 = 0$. $k^2 = 1$. Thus,$k = \pm 1$.

Class 11 Mathematics Pair of straight lines Equation of lines joining the origin to the point of intersection of a curve and a line and Distance between the pair of lines

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The line $x+2y=k$ meets the curve $2x^2-2xy+3y^2+2x-y-1=0$ at two points $A$ and $B$. Let $O$ be the origin. If the line segments $OA$ and $OB$ are perpendicular to each other,then $k=$

TS EAMCET 2022 All TS EAMCET

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$\pm 1$
B
$\pm 2$
C
$\pm 3$
D
$4$

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Equation of lines joining the origin to the point of intersection of a curve and a line and Distance between the pair of lines

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The line $x+2y=k$ meets the curve $2x^2-2xy+3y^2+2x-y-1=0$ at two points $A$ and $B$. Let $O$ be the origin. If the line segments $OA$ and $OB$ are perpendicular to each other,then $k=$

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