Question 131018
for line a, first you need to find out the slope of line a.

slope of line a = (5-3)/(5-3) = 2/2=1 = m 

use line equation y -y1 = m(x-x1) where (x1,y1) is either your (3,3) or (5,5).

so the equation of line a is:

y-3 = 1*(x-3)

y-3 = x-3

y = x --> line a

for line b, first you need to find out the slope of line b.

slope of line a = (-5+3)/(5-3) = -2/2= -1 = m 

use line equation y -y1 = m(x-x1) where (x1,y1) is either your (3,-3) or (5,-5).

so the equation of line a is:

y+3 = -1*(x-3)

y+3 = -x+3

y = -x --> line b

now, to find out the intersection point of these two lines.

y=x (line a) and y=-x (line b)

subtitute y=x into equation of line b,

so we have y = -y , Hence, y=0.

now subtitude y=0 to equation of line a . we have x=0

so (0,0) is the intersection point of line a and b.