Let ABCD be the rectangle window with A(0, 0), B(10, 0), C(10, 10) and D(0, 10). Use Liang Barsky line clipping algorithm to clip the line XY where X(5, 3) and Y(15, 9)

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Suresh Chand · Accepted Answer

Solution: xwmin = 0 ywmin = 0 xwmax = 10 ywmax = 10 (x1, y1) = (5, 3) (x2, y2) = (15, 9) Δx = x2 - x1 = 15 - 5 = 10 Δy = y2 - y1 = 9 - 3 = 6 P1 = -Δx = -10 P2 = Δx = 10 P3 = -Δy = -6 P4 = Δy = 6 q1 = x1 - xwmin = 5 q2 = xwmax - x1 = 5 q3 = y1 - ywmin = 3 q4 = ywmax - y1 = 7 Since, The value of P is not 0. So, the line is not parallel to the view. Pk < 0 P1 and P3 Now we will find tmax (t_{max} = maxleft ( 0, frac{q_k}{p_k} right )) (= maxleft ( 0, frac{5}{-10}, frac{3}{-6} right )) = 0 Now, We will find coordinates using below formula x = x1 + tΔx y = y1 + Δy x = 5 + 0.10 = 5 y = 3 + 0.6 = 3 (x, y) = (5, 3)  Pk > 0 P2,  P4 Now we will find tmin (t_{min} = minleft ( 1, frac{q_k}{p_k} right )) (= minleft ( 1, frac{5}{10}, frac{7}{6} right )) (= frac{1}{2}) Now, We will find coordinates using below formula x = x1 + tΔx y = y1 + Δy (x = 5 + frac{1}{2} . 10 = 10) (y = 3 + frac{1}{2} . 6 = 6) (x, y) = 10, 6) So, the final line after clipping is given below

Let ABCD be the rectangle window with A(0, 0), B(10, 0), C(10, 10) and D(0, 10). Use Liang Barsky line clipping algorithm to clip the line XY where X(5, 3) and Y(15, 9)

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Let ABCD be the rectangle window with A(0, 0), B(10, 0), C(10, 10) and D(0, 10). Use Liang Barsky line clipping algorithm to clip the line XY where X(5, 3) and Y(15, 9)

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