(a) y-intercept and slope: In a linear equation \(\displaystyle{y}={b}_{{0}}+{b}_{{1}}{x}\), the constant b1 be the slope and b0 be the y-intercept form and x is the independent variable and y is the dependent variable.

Comparing the given equation with the general form of linear equation the slope of the equation is 0 and the y-intercept is –3. Thus, the slope of the linear equation is \(\displaystyle{b}_{{1}}={0}\) and the y-intercept is \(\displaystyle{b}_{{0}}=-{3}\)

(b) It is known that, the slope of the linear equation \(\displaystyle{y}={b}_{{0}}+{b}_{{1}}{x}\) is upward if \(\displaystyle{b}_{{1}}{>}{0}\), the slope of the linear equation \(\displaystyle{y}={b}_{{0}}+{b}_{{1}}{x}\) is downward if \(\displaystyle{b}_{{1}}{<}{0}\), and the slope of the linear equation \(\displaystyle{y}={b}_{{0}}+{b}_{{1}}{x}\) is horizontal if \(\displaystyle{b}_{{1}}={0}\) .

Thus, in the given equation \(\displaystyle{y}=-{3}{b}_{{1}}={0}\)

Thus, the slope is horizontal.

(c) The two points \(\displaystyle{\left({x}_{{1}},{y}_{{1}}\right)}{\quad\text{and}\quad}{\left({x}_{{2}},{y}_{{2}}\right)}\) on the given line are obtained

If \(x=0\)

\(y=-3\)

Thus, one point on the line is \(\displaystyle{\left({x}_{{1}},{y}_{{1}}\right)}={\left({0},-{3}\right)}.\)

If \(x=1\)

\(y=-3\)

Thus, the second point on the line is \(\displaystyle{\left({x}_{{2}},{y}_{{2}}\right)}={\left({1},-{3}\right)}\)