Algebra Know-It-ALL

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.

8. 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.


9. 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.