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The pairwise intersections of the three lines x - y + 1 = 0, x - 2y + 3 = 0, and 2x - 5y + 11 = 0 form three points that are the midpoints of the sides of a triangle ABC. What is the area of triangle ABC?
- 6
- 8
- 10
- 12
Correct answer: 6
Solution
Find three intersection points: P1 (L1∩L2), P2 (L1∩L3), P3 (L2∩L3). P1: x-y+1=0 and x-2y+3=0. Subtracting: y-2=0 -> y=2, x=1. P1=(1,2). P2: x-y+1=0 and 2x-5y+11=0. From L1: x=y-1. Sub: 2(y-1)-5y+11=0 -> 2y-2-5y+11=0 -> -3y+9=0 -> y=3, x=2. P2=(2,3). P3: x-2y+3=0 and 2x-5y+11=0. From L1: x=2y-3. Sub: 2(2y-3)-5y+11=0 -> 4y-6-5y+11=0 -> -y+5=0 -> y=5, x=7. P3=(7,5). Area of medial triangle (P1P2P3): using formula = (1/2)|det([P2-P1, P3-P1])| = (1/2)|det([(1,1),(6,3)])| = (1/2)|3-6| = (1/2)*3 = 3/2. Area of ABC = 4 * (3/2) = 6.
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