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homework 1 solutions

# homework 1 solutions - CSE 152 Homework 1 solution Question...

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CSE 152 – Homework 1 solution Question 1 a. In Euclidean coordinates, denote the line as y = ax + b, we have 1 = a + b pass through (1,1) 2 = 3a + b pass through (3,2) => a = 1/2, b = 1/2 => y = x/2 + 1/2 => -x + 2y -1 = 0 b. In homogeneous coordinates, the line connecting (1,1) and (3,2) is (1,1,1) × (3,2,1) = (-1,2,-1) => line equation: -x + 2y -1 = 0 c. Four corners: A=(1,3), B=(2,4), C=(5,2), D=(2,1) => 4 edges AB: (1,3,1) × (2,4,1) = (-1,1,-2) => -x +y -2 = 0 BC: (2,4,1) × (5,2,1) = (2,3,-16) => 2x + 3y -16 = 0 CD: (5,2,1) × (2,1,1) = (1,-3,1) => x – 3y +1 = 0 DA: (2,1,1) × (1,3,1) = (-2,-1,5) => -2x -y + 5 = 0 Two vanishing points AB × CD = (-1,1,-2) × (1,-3,1) = (-5,-1,2) => (-5/2, -1/2) in Euclidean BC × DA = (2,3,-16) × (-2,-1,5) = (-1,22,4) => (-1/4, 22/4) in Euclidean Question 3 Homogeneous coordinates of the rectangle (4 corners): P 1 = (-6, -3, 6, 1) T , P 2 = (3, -3, 6, 1) T , P 3 = (3, 3, 9, 1) T , P 4 = (-6, 3, 9, 1) T a. = 0 3 1 0 0 0 0 1 0 0 0 0 1 e perspectiv M . The projection of the rectangle: M perspective * P 1 = (-6, -3, 2) T => Euclidean (-3, -1.5) T M perspective * P 2 = (3, -3, 2) T => Euclidean (1.5, -1.5) T M perspective * P 3 = (3, 3, 3) T => Euclidean (1, 1) T M perspective * P 4 = (-6, 3, 3) T => Euclidean (-2, 1) T b.

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