To find the parametric equation of the line with slope -2 passing through (-10,-20)

Given slope \(\displaystyle{m}={\frac{{\triangle{y}}}{{\triangle{x}}}}={\frac{{{y}_{{{s}}}}}{{{x}_{{{s}}}}}}=-{2}\)

\(\displaystyle\Rightarrow{x}_{{{s}}}={1},{y}_{{{s}}}=-{2}\)

Also \(\displaystyle{\left({x}_{{{1}}},{y}_{{{1}}}\right)}={\left(-{10},-{20}\right)}\)

The parametric equations of the line is given by

\(\displaystyle{x}={x}_{{{1}}}+{x}_{{{s}}}\cdot{t}\)

\(\displaystyle{y}={y}_{{{1}}}+{y}_{{{s}}}\cdot{t}\)

Substituting the values we have

\(\displaystyle{x}={x}_{{{1}}}+{x}_{{{s}}}\cdot{t}=-{10}+{1}{t}={t}-{10}\)

\(\displaystyle{y}={y}_{{{1}}}+{y}_{{{s}}}\cdot{t}=--{20}-{2}{t}=-{2}{t}-{20}\)

Hence x=t-10, y=-2t-20 are the required parametric equations.

Given slope \(\displaystyle{m}={\frac{{\triangle{y}}}{{\triangle{x}}}}={\frac{{{y}_{{{s}}}}}{{{x}_{{{s}}}}}}=-{2}\)

\(\displaystyle\Rightarrow{x}_{{{s}}}={1},{y}_{{{s}}}=-{2}\)

Also \(\displaystyle{\left({x}_{{{1}}},{y}_{{{1}}}\right)}={\left(-{10},-{20}\right)}\)

The parametric equations of the line is given by

\(\displaystyle{x}={x}_{{{1}}}+{x}_{{{s}}}\cdot{t}\)

\(\displaystyle{y}={y}_{{{1}}}+{y}_{{{s}}}\cdot{t}\)

Substituting the values we have

\(\displaystyle{x}={x}_{{{1}}}+{x}_{{{s}}}\cdot{t}=-{10}+{1}{t}={t}-{10}\)

\(\displaystyle{y}={y}_{{{1}}}+{y}_{{{s}}}\cdot{t}=--{20}-{2}{t}=-{2}{t}-{20}\)

Hence x=t-10, y=-2t-20 are the required parametric equations.