The lines x+y+4=0,x-2y-4=0,and 3x+4y-2=0: | vedclass

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(D) To check if the lines are concurrent,we solve the first two equations for their point of intersection: $x+y = -4$ $(1)$ $x-2y = 4$ $(2)$ Subtracti

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form a scalene triangle

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(D) To check if the lines are concurrent,we solve the first two equations for their point of intersection: $x+y = -4$ $(1)$ $x-2y = 4$ $(2)$ Subtracting $(2)$ from $(1)$: $(x+y) - (x-2y) = -4 - 4 \implies 3y = -8 \implies y = -8/3$. Substituting $y = -8/3$ into $(1)$: $x - 8/3 = -4 \implies x = -4 + 8/3 = -4/3$. The intersection point is $(-4/3, -8/3)$. Now,check if this point satisfies the third equation $3x+4y-2=0$: $3(-4/3) + 4(-8/3) - 2 = -4 - 32/3 - 2 = -6 - 32/3 = -18/3 - 32/3 = -50/3 
eq 0$. Since the point does not satisfy the third equation,the lines are not concurrent. Next,find the slopes of the lines: $L_1: x+y+4=0 \implies m_1 = -1$. $L_2: x-2y-4=0 \implies m_2 = 1/2$. $L_3: 3x+4y-2=0 \implies m_3 = -3/4$. Since no two slopes are equal,the lines are not parallel. Since $m_1 	imes m_2 = -1/2 
eq -1$,$m_2 	imes m_3 = -3/8 
eq -1$,and $m_1 	imes m_3 = 3/4 
eq -1$,no two lines are perpendicular. Calculate the lengths of the sides of the triangle formed by the intersection points: $P_1 (L_1 \cap L_2) = (-4/3, -8/3)$. $P_2 (L_2 \cap L_3)$: $x-2y=4$ and $3x+4y=2$. Solving gives $x=2, y=-1$. $P_2 = (2, -1)$. $P_3 (L_3 \cap L_1)$: $3x+4y=2$ and $x+y=-4$. Solving gives $x=18, y=-22$. $P_3 = (18, -22)$. Calculate side lengths: $d_{12} = \sqrt{(2 - (-4/3))^2 + (-1 - (-8/3))^2} = \sqrt{(10/3)^2 + (5/3)^2} = \sqrt{125/9} = 5\sqrt{5}/3$. $d_{23} = \sqrt{(18 - 2)^2 + (-22 - (-1))^2} = \sqrt{16^2 + (-21)^2} = \sqrt{256 + 441} = \sqrt{697}$. $d_{31} = \sqrt{(18 - (-4/3))^2 + (-22 - (-8/3))^2} = \sqrt{(58/3)^2 + (-58/3)^2} = 58\sqrt{2}/3$. Since all side lengths are distinct,the triangle is scalene.

The lines $x+y+4=0$,$x-2y-4=0$,and $3x+4y-2=0$:

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