Line L_1 with slope 2 and line L_2 with slope | vedclass

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(B) The total number of points is $1 + 12 + 9 = 22$. To form a triangle,we need to select $3$ non-collinear points. The points $O, P_1, \ldots, P_{12}

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(B) The total number of points is $1 + 12 + 9 = 22$. To form a triangle,we need to select $3$ non-collinear points. The points $O, P_1, \ldots, P_{12}$ are collinear on $L_1$,and $O, Q_1, \ldots, Q_9$ are collinear on $L_2$. Total ways to select $3$ points from $22$ is $\binom{22}{3} = \frac{22 	imes 21 	imes 20}{3 	imes 2 	imes 1} = 1540$. We must subtract the cases where the $3$ points are collinear: $1$. $3$ points on $L_1$: $\binom{13}{3} = \frac{13 	imes 12 	imes 11}{6} = 286$. $2$. $3$ points on $L_2$: $\binom{10}{3} = \frac{10 	imes 9 	imes 8}{6} = 120$. Total triangles = $1540 - 286 - 120 = 1134$.

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