Question

write the equation of a line in standard form with a slope of -3 and passes through the point (3,-8).
-3x - y = 1
3x - y = 1
3x - y = -1
3x + y = 1

Explanation:

Step1: Use point - slope form

The point - slope form of a line is $y - y_1=m(x - x_1)$, where $m$ is the slope and $(x_1,y_1)$ is a point on the line. Given $m=-3$ and the point $(3,-8)$, we substitute these values into the point - slope form:
$y-(-8)=-3(x - 3)$
Simplify the left - hand side: $y + 8=-3(x - 3)$

Step2: Expand the right - hand side

Using the distributive property $a(b - c)=ab - ac$, we have:
$y + 8=-3x+9$

Step3: Convert to standard form

The standard form of a line is $Ax+By = C$, where $A$, $B$, and $C$ are integers and $A\geq0$.
We want to get all the $x$ and $y$ terms on one side and the constant on the other. Add $3x$ to both sides of the equation $y + 8=-3x + 9$:
$3x+y+8 = 9$
Then subtract 8 from both sides:
$3x + y=9 - 8$
$3x + y=1$

Answer:

$3x + y = 1$ (corresponding to the option "3x + y = 1")