The two pairs of straight lines 12x^2+7xy-12y | vedclass

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(A) Factorizing $12x^2+7xy-12y^2=0$:$12x^2+16xy-9xy-12y^2=0$$4x(3x+4y)-3y(3x+4y)=0$$(4x-3y)(3x+4y)=0$So,the lines are $4x-3y=0$ and $3x+4y=0$. Since t

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area of square $\frac{1}{25}$ sq units

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(A) Factorizing $12x^2+7xy-12y^2=0$:$12x^2+16xy-9xy-12y^2=0$$4x(3x+4y)-3y(3x+4y)=0$$(4x-3y)(3x+4y)=0$So,the lines are $4x-3y=0$ and $3x+4y=0$. Since their slopes are $4/3$ and $-3/4$,they are perpendicular.Factorizing $12x^2+7xy-12y^2-x+7y-1=0$:Let the equation be $(4x-3y+c_1)(3x+4y+c_2)=0$.Expanding this: $12x^2+7xy-12y^2 + (4c_2+3c_1)x + (4c_1-3c_2)y + c_1c_2 = 0$.Comparing with $12x^2+7xy-12y^2-x+7y-1=0$:$4c_2+3c_1 = -1$$4c_1-3c_2 = 7$Solving these,we get $c_1=1$ and $c_2=-1$.So the lines are $4x-3y+1=0$ and $3x+4y-1=0$.The distance between parallel lines $4x-3y=0$ and $4x-3y+1=0$ is $d_1 = \frac{|1-0|}{\sqrt{4^2+(-3)^2}} = \frac{1}{5}$.The distance between parallel lines $3x+4y=0$ and $3x+4y-1=0$ is $d_2 = \frac{|-1-0|}{\sqrt{3^2+4^2}} = \frac{1}{5}$.Since the lines are perpendicular and the distance between parallel pairs is equal,the figure is a square with side length $1/5$.Area $= (1/5)^2 = 1/25$ sq units.

The two pairs of straight lines $12x^2+7xy-12y^2=0$ and $12x^2+7xy-12y^2-x+7y-1=0$ constitute a

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