Solution:
The equation of a line passing through two given points is expressed as
[tex]\begin{gathered} y-y_1=(\frac{y_2-y_1}{x_2-x_1})(x-x_1)\text{ ----- equation 1} \\ where \\ (x_1,y_1)\text{ and \lparen x}_2,y_2)\text{ are the points through which the line passes} \end{gathered}[/tex]Given that the line passes through the points (20,-14) and (12, -12), this implies that
[tex]\begin{gathered} x_1=20 \\ y_1=-14 \\ x_2=12 \\ y_2=-12 \end{gathered}[/tex]thus, the equation of the line becomes
[tex]\begin{gathered} y-(-14)=)=(\frac{-12-(-14)}{12-20})(x-20) \\ thus, \\ y+14=(\frac{-12+14}{12-20})(x-20) \\ y+14=-\frac{1}{4}(x-20) \\ open\text{ parentheses,} \\ y+14=-\frac{1}{4}x+\frac{20}{4} \\ subtract\text{ 14 from both sides of the equation} \\ y+14-14=-\frac{1}{4}x+5-14 \\ \Rightarrow y=-\frac{1}{4}x-11 \end{gathered}[/tex][tex]\begin{gathered} But, \\ \frac{dy}{dx}=\frac{dy}{dt}\times\frac{dt}{dx} \end{gathered}[/tex]Thus,
[tex]\begin{gathered} Given\text{ that} \\ y=t-5, \\ \frac{dy}{dt}=1 \\ Also, \\ y=-\frac{1}{4}x-11 \\ \frac{dy}{dx}=-\frac{1}{4} \\ thus, \\ -\frac{1}{4}=1\times\frac{dt}{dx} \\ \Rightarrow\frac{dx}{dt}=-4 \end{gathered}[/tex]By integration,
[tex]\begin{gathered} \frac{dx}{dt}=-4 \\ \Rightarrow\int dx=-4\int dt \\ thus, \\ x=-4t+c \end{gathered}[/tex]