The number of triangles formed by the lines x | vedclass

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(A) Let the lines be $L_1: x-y+3=0$,$L_2: 2x-y+3=0$,$L_3: 3x-y+2=0$,and $L_4: x+y-3=0$.To form a triangle,we need $3$ lines that are not concurrent an

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$4$

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(A) Let the lines be $L_1: x-y+3=0$,$L_2: 2x-y+3=0$,$L_3: 3x-y+2=0$,and $L_4: x+y-3=0$.To form a triangle,we need $3$ lines that are not concurrent and not parallel.First,check for concurrency by solving the system of equations for any $3$ lines.Solving $L_1, L_2, L_3$: The intersection of $L_1$ and $L_2$ is $(-0, 3)$. Substituting into $L_3$: $3(0)-3+2 = -1 
eq 0$. Thus,they are not concurrent.Checking all combinations of $3$ lines out of $4$ ($^4C_3 = 4$ combinations):$1$. $(L_1, L_2, L_3)$: Slopes are $1, 2, 3$. No two lines are parallel. They form a triangle.$2$. $(L_1, L_2, L_4)$: Slopes are $1, 2, -1$. No two lines are parallel. They form a triangle.$3$. $(L_1, L_3, L_4)$: Slopes are $1, 3, -1$. No two lines are parallel. They form a triangle.$4$. $(L_2, L_3, L_4)$: Slopes are $2, 3, -1$. No two lines are parallel. They form a triangle.Since no three lines are concurrent,the total number of triangles formed is $4$.

The number of triangles formed by the lines $x-y+3=0$,$2x-y+3=0$,$3x-y+2=0$,and $x+y-3=0$ is:

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