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

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(A) The first pair of lines is $12x^2+7xy-12y^2=0$. Factoring this,we get $12x^2+16xy-9xy-12y^2=0$,which simplifies to $(4x-3y)(3x+4y)=0$. Thus,the li

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

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(A) The first pair of lines is $12x^2+7xy-12y^2=0$. Factoring this,we get $12x^2+16xy-9xy-12y^2=0$,which simplifies to $(4x-3y)(3x+4y)=0$. Thus,the lines are $4x-3y=0$ and $3x+4y=0$. Their slopes are $m_1 = \frac{4}{3}$ and $m_2 = -\frac{3}{4}$. Since $m_1 	imes m_2 = -1$,these lines are perpendicular.The second pair of lines is $12x^2+7xy-12y^2-x+7y-1=0$. Factoring this quadratic in $x$,we solve for $x$ using the quadratic formula: $x = \frac{-(7y-1) \pm \sqrt{(7y-1)^2 - 4(12)(-12y^2+7y-1)}}{24}$. Simplifying the discriminant gives $\sqrt{625y^2-350y+49} = \sqrt{(25y-7)^2} = 25y-7$. This yields the lines $24x-18y+6=0$ (or $4x-3y+1=0$) and $24x+32y-8=0$ (or $3x+4y-1=0$).The lines $4x-3y=0$ and $4x-3y+1=0$ are parallel,and $3x+4y=0$ and $3x+4y-1=0$ are parallel. The distance between the 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 $3x+4y=0$ 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 distances between parallel pairs are equal,the figure is a square with side length $\frac{1}{5}$. The area is $(\frac{1}{5})^2 = \frac{1}{25}$ sq units.

$x$	$0$	$1$	$2$	$3$	$4$	$5$
$f(x)$	$2$	$3$	$6$	$11$	$18$	$27$

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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