The equation of the pair of straight lines pa | vedclass

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(B) The given pair of lines is $3x^2-8xy+5y^2=0$.The equation of the pair of lines passing through the origin and perpendicular to the given lines is

Vedclass · Accepted Answer

$5x^2+8xy+3y^2-18x-14y+16=0$

Vedclass · Answer

(B) The given pair of lines is $3x^2-8xy+5y^2=0$.The equation of the pair of lines passing through the origin and perpendicular to the given lines is obtained by interchanging the coefficients of $x^2$ and $y^2$ and changing the sign of the $xy$ term: $5x^2+8xy+3y^2=0$.To find the equation of the pair of lines passing through $(1,1)$,we replace $x$ with $(x-1)$ and $y$ with $(y-1)$ in the equation $5x^2+8xy+3y^2=0$.This gives $5(x-1)^2+8(x-1)(y-1)+3(y-1)^2=0$.Expanding this,we get $5(x^2-2x+1)+8(xy-x-y+1)+3(y^2-2y+1)=0$.Simplifying,$5x^2-10x+5+8xy-8x-8y+8+3y^2-6y+3=0$.Combining like terms,$5x^2+8xy+3y^2-18x-14y+16=0$.

The equation of the pair of straight lines passing through the point $(1,1)$ and perpendicular to the pair of straight lines $3x^2-8xy+5y^2=0$ is

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