Solution:

No. of points = 4 = k

Degree of curve $=(\mathrm{n})=\mathrm{k}-1=4-1=3$

$\therefore$ Cubic Curve

$P(t)=P_{0}(1-t)^{3}+3 P_{1}(1-t)^{2} t+3 P_{2}(1-t) t^{2}+P_{3} t^{3}$

$P_{0}=\left[\begin{array}{ll}{20} & {20}\end{array}\right]$

$P_{1}=\left[\begin{array}{ll}{60} & {80}\end{array}\right]$

$P_{2}=\left[\begin{array}{ll}{120} & {100}\end{array}\right]$

$P_{3}=\left[\begin{array}{ll}{150} & {30}\end{array}\right]$

$\begin{aligned} P_{x} &=20(1-t)^{3}+3(60)(1-t)^{2} t+3(120)(1-t) t^{2}+150 t^{3} \\ &=20\left(1-3 t+3 t^{2}-t^{3}\right)+180\left(1-2 t+t^{2}\right) t+360\left(t^{2}-t^{3}\right)+150 t^{3} \end{aligned}$

$=20-60 t+60 t^{2}-20 t^{3}+180 t-360 t^{2}+180 t^{3}+360 t^{2}-360 t^{3}+150 t^{3}$

$\therefore P_{x}=20+120 t+60 t^{2}-50 t^{3}$

$\begin{aligned} P_{y} &=20\left(1-3 t+3 t^{2}-t^{3}\right)+240\left(1-2 t+t^{2}\right) t+300\left(t^{2}-t^{3}\right)+30 t^{3} \\ &=20-60 t+60 t^{2}-20 t^{3}+240 t-480 t^{2}+240 t^{3}+300 t^{2}-300 t^{3}+30 t^{3} \end{aligned}$

$\therefore P_{y}=20+180 t-120 t^{2}-50 t^{3}$

For Mid point, $\mathrm{t}=0.5$

$P_{x}=88.75$

$P_{y}=73.75$

Slope $=\frac{d_{y}}{d_{x}}$

$\frac{d_{y}}{d_{x}}=\frac{0+120+120 t-150 t^{2}}{0+180-240 t-150 t^{2}}$

Slope $=\frac{d y}{d_{x}}=\frac{\frac{d y}{d_{t}}}{d_{t}}=\frac{180-240 t-150 t^{2}}{120+120 t-150 t^{2}}$

$\therefore$ Slope $=0.15789$

Parametric Representation of a Line:

$P(t)=C_{0}+C_{1} t$

At start point $P(t)=P_{0}, t=0$

$\therefore P_{0}=C_{0}$

At End point $P(t)=P_{1}, t=1$

$\therefore C_{1}=P_{1}-P_{0}$

$P_{x}=x_{0}+\left(x_{1}-x_{0}\right) t$

$P_{y}=y_{0}+\left(y_{1}-y_{0}\right) t$