The line frac{x}{3}+frac{y}{2}=1 and a pair o | vedclass

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(B) The given line is $\frac{x}{3}+\frac{y}{2}=1$,which can be written as $2x+3y=6$,or $y=-\frac{2}{3}x+2$. The slope of this line is $m_1=-\frac{2}{3

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$5\left(x^2-y^2\right)-24 x y=0$

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(B) The given line is $\frac{x}{3}+\frac{y}{2}=1$,which can be written as $2x+3y=6$,or $y=-\frac{2}{3}x+2$. The slope of this line is $m_1=-\frac{2}{3}$.Since the pair of lines passing through the origin are perpendicular and form an isosceles triangle with the given line,the angles they make with the given line must be $45^{\circ}$.Let the slopes of the pair of lines be $m$ and $-\frac{1}{m}$ (since they are perpendicular).The angle between the line $y=mx$ and the line $y=-\frac{2}{3}x+2$ is $45^{\circ}$.$	an 45^{\circ} = \left| \frac{m - (-2/3)}{1 + m(-2/3)} ight| = 1$$1 = \left| \frac{3m+2}{3-2m} ight|$$3m+2 = 3-2m$ or $3m+2 = -(3-2m)$$5m = 1 \Rightarrow m = \frac{1}{5}$ or $m+5 = 0 \Rightarrow m = -5$.The lines are $y=\frac{1}{5}x$ and $y=-5x$,which are $x-5y=0$ and $5x+y=0$.The joint equation is $(x-5y)(5x+y) = 0$.$5x^2 + xy - 25xy - 5y^2 = 0$$5x^2 - 24xy - 5y^2 = 0$$5(x^2-y^2) - 24xy = 0$.

The line $\frac{x}{3}+\frac{y}{2}=1$ and a pair of lines both passing through the origin form an isosceles triangle. If this pair of lines are perpendicular,then the equation of the pair of straight lines is

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