636 Worked-Out Solutions to Exercises: Chapters 11 to 19
Dividing through by 2, we get t= 5. That’s the t value of the intersection point. We can plug
5 in for t in either of the original equations to solve for s. Let’s use the first one. We then get
5 =s+ 5
It’s not too difficult to tell from this equation that s= 0. Now we know that s= 0 and
t= 5, so (s,t)= (0, 5) defines the point where the two lines intersect. That point lies on
thet axis, because the s coordinate is equal to 0.
- Figure B-7 shows the graphs of the two lines, based on their known slopes and
t-intercepts. The intersection point is, coincidentally, at the t-intercept for both lines.
The lines intersect at a 90° angle because the axes are graduated in equal increments. - For reference, here’s the general two-point equation we derived for a line in Cartesian
coordinates:
y−y 1 = (x−x 1 )(y 2 −y 1 ) / (x 2 −x 1 )
where points are represented by ordered pairs (x 1 ,y 1 ) and (x 2 ,y 2 ). We are told that (2, 8)
and (0, −4) both lie on the graph. Let’s assign x 1 = 2, x 2 = 0, y 1 = 8, and y 2 =−4. When
we plug these numbers into the above equation, we get
y− 8 = (x− 2)(− 4 − 8) / (0 − 2)
s
t
t=s+ 5
t=–s+ 5
Intersection
point = (0,5)
10
15
–5
–10
–15
–10–15 –5^51015
Figure B-7 Illustration for the solution to Prob. 8 in
Chap. 15.