Question

Interactive computer graphics course.

by java language, no need to explain.

Q1: Draw a line between P0 and Pe

P0= (x,y)=(2,5)

Pe=(xe,ye)=(7,9)

Q2: P=(3,5) Scale P by (5,7) and then sheer by -3 along y-axis

Answer 1

Ans a) P0=(x,y)=(2,5)

Pe=(xe,ye)=(7,9)

Parametric equation of line is y=mx+b

where m=(ye-ys)/(xe-xs)

m=(9-5)/(7-2)=4/5=0.8

m<1 indicates that line is closer to x

then x_i+1=x_i+1

and y_i+1=y_i+m at each step of DDA algorithm

x	y	plotted y
2	5	2,5
2+1=3	5+0.8=5.8	3,6
3+1=4	5.8+0.8=6.6	4,7
4+1=5	6.6+0.8=7.4	5,7
5+1=6	7.4+0.8=8.2	6,8

6+1=7

8.2+0.8=9.0

7,9

now we have reached the point 7,9 so we will stop here and according to the plotted pts line can be drawn(we always use integer value for plotting a line in dda algorithm so pt y is rounded off for plotting).

Ans 2)P(x,y)=3,5

scaling factor Sx=5 and Sy=7 are given

After scaling

P(x')=Sx.x=5*3=15

P(y')=sy.y=7*5=35

So after scaling the point will be P(15,35)

Now On shearing this pt by -3 along y axis

means

now we have x=15,y=35

formula for shearing along y axis is

y'=y+shy.x and x'=x

where shy is shearing value along y axis

which is given as -3

now using the formula we get

x'=x=15

y'=35+15*(-3)=35-45=-10

So after shearing new coordinates will be 15,-10

So answer will be P(15,-10)