Question

suppose parametric equations for the line segment between (4,9) and (1,6) have the form: \

$$\begin{cases} x(t) = a + bt \\\\ y(t) = c + dt \\end{cases}$$

if the parametric curve starts at (4,9) when t = 0 and ends at (1,6) at t = 1, then find a, b, c, and d. a = \boxed{} b = \boxed{} c = \boxed{} d = \boxed{} question help: \boxed{\text{video}}

Explanation:

Step1: Find \(a\) using \(t = 0\) in \(x(t)\)

When \(t = 0\), \(x(0)=a + b\times0=a\). And \(x(0) = 4\) (since the curve starts at \((4,9)\)), so \(a = 4\).

Step2: Find \(b\) using \(t = 1\) in \(x(t)\)

When \(t = 1\), \(x(1)=a + b\times1=a + b\). We know \(x(1)=1\) and \(a = 4\), so \(1=4 + b\). Solving for \(b\), we get \(b=1 - 4=- 3\).

Step3: Find \(c\) using \(t = 0\) in \(y(t)\)

When \(t = 0\), \(y(0)=c + d\times0=c\). And \(y(0) = 9\) (since the curve starts at \((4,9)\)), so \(c = 9\).

Step4: Find \(d\) using \(t = 1\) in \(y(t)\)

When \(t = 1\), \(y(1)=c + d\times1=c + d\). We know \(y(1)=6\) and \(c = 9\), so \(6=9 + d\). Solving for \(d\), we get \(d=6 - 9=-3\).

Answer:

\(a = 4\), \(b=-3\), \(c = 9\), \(d=-3\)