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Find the area (in square units) of triangle ABC where A = (9, -9), B = (1, 3), and C lies on the line 3x + 2y + 4 = 0.
- sqrt(13)
- 26
- sqrt(26)
- sqrt(208)
Correct answer: 26
Solution
Direction of AB: B-A = (1-9, 3-(-9)) = (-8, 12). Slope of AB = 12/(-8) = -3/2. The line 3x+2y+4=0 has slope = -3/2. YES — AB is parallel to 3x+2y+4=0! So C's perpendicular distance to AB is constant (same as the distance between the two parallel lines), and the area of triangle ABC is fixed. Line AB: slope -3/2 through A(9,-9): y+9 = -3/2*(x-9) => 2y+18 = -3x+27 => 3x+2y-9=0. Distance between lines 3x+2y-9=0 and 3x+2y+4=0: d = |(-9)-(4)|/sqrt(9+4) = |-13|/sqrt(13) = 13/sqrt(13) = sqrt(13). |AB| = sqrt((-8)²+12²) = sqrt(64+144) = sqrt(208) = 4*sqrt(13). Area = (1/2)*|AB|*d = (1/2)*4*sqrt(13)*sqrt(13) = (1/2)*4*13 = 26.
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