Show that the graph of the parametric equations

$x=a+(c-a)t,y=b+(d-b)t,t\in [0,1]$

is a line segment from $(a,b)$to$(c,d).$

The graph of the parametric equations

$x=a+(c-a)t,y=b+(d-b)t,t\in [0,1]$

is a line segment from $(a,b)$to

Consider the two points $(a,b)(c,d)$.

The objective is to show that the line segment joining the two points is equal to the graph of the parametric equations.

If $\left(x_1,y_1\right)\left(x_2,y_2\right)$are any two points, then the equation for the line segment is given by $x=(1-t)x_1+t{x}_2,y=(1-t)y_1+t{y}_2$for some real number$t$.

Now write the line segment equations for the points localid="1653036356110" $(a,b)(c,d)$.

Then,

localid="1653036358431" $$\begin{gathered} x=(1-t)a+tc\left[since{x}_1=a,x_2=c\right] \\ x=a-ta+tc \end{gathered}$$

By adjusting the terms

Now,

localid="1653036360951" $$\begin{gathered} y=(1-t)b+td\left[\text{since}{y}_1=b_{,}{y}_2=d\right] \\ y=b-tb+td \end{gathered}$$

localid="1653036364252" $y=b+(d-b)t$By adjusting the terms

Thus the line segment joining the points, localid="1653036370197" $(a,b)(c,d)$is localid="1653036367612" $x=a+(c-a)t,y=b+(d-b)t$which is equal to the given parametric equations.

Hence proved.