Question

Suppose parametric equations for the line segment between (9,0) and (−2,10) have the form: x=a+bt and y=c+dt

If the parametric curve starts at (9,7) when t=0 and ends at (5,-2) at t=1 , then find a,b,c and d.

Answer 1

a= 9

b= -4

c= 7

d= -9

x = 9 - 4t

y = 7 - 9t

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Explanation:

When t = 0, the parametric curve starts at (9,7). Substituting these values into the parametric equations, we get:

x = a + bt becomes 9 = a + b(0), so a = 9.

y = c + dt becomes 7 = c + d(0), so c = 7.

When t = 1, the parametric curve ends at (5,-2). Substituting these values into the parametric equations, we get:

x = a + bt becomes 5 = a + b(1), so b = 5 - a.

y = c + dt becomes -2 = c + d(1), so d = -2 - c.

Substituting the values of a and c that we found earlier, we get:

b = 5 - a becomes b = 5 - 9, so b = -4.

d = -2 - c becomes d = -2 - 7, so d = -9.

So the parametric equations for the line segment between (9,7) and (5,-2) are:

x = 9 - 4t

y = 7 - 9t