What is the equation of the line that passes through the point (3,-6) and has a slope of −3?

A line has only one dimension i.e., length and no width.
It is formed with an infinite number of points lying on it.
It can be extended infinitely along a straight path in both directions.
An infinite number of lines pass through a given point, and a unique straight line can be drawn if any two points on that line are shown.

The slope of a line:

The slope of a line is the tangent value of the angle [tex]\theta[/tex] i.e., [tex]\tan \theta[/tex], where [tex]\theta[/tex] is the angle made by the line with the positive [tex]x[/tex]-axis.

Equation of a line:

The equation of a line in slope-intercept form is given by [tex]y=mx+c[/tex], where [tex]m[/tex] is the slope of the line and [tex]c[/tex] is the [tex]y[/tex] intercept of the line.

Here, we want to find the equation of the line which passes through the point [tex](3,-6)[/tex] and has a slope [tex]-3[/tex].

So, put [tex]m=-3[/tex] in [tex]y=mx+c[/tex] and obtain [tex]y=-3x+c[/tex].

Now, since the line passes through the point [tex](3,-6)[/tex], put [tex]x=3[/tex] and [tex]y=-6[/tex] in [tex]y=-3x+c[/tex] to obtain:

[tex]-6=-3\times 3+c\\\Longrightarrow -6=-9+c\\\Longrightarrow c=-6+9\\\Longrightarrow c=3[/tex]

Therefore, after substituting [tex]c=3[/tex] in [tex]y=-3x+c[/tex], we obtain the required equation of the line that is [tex]y=-3x+3[/tex].

So, the equation of the line that passes through the point [tex](3,-6)[/tex] and has a slope [tex]-3[/tex] is [tex]y=-3x+3[/tex].