Question

a line passes through point (5, -6) and has a slope of \frac{2}{3}. write an equation in standard form for this line.

Explanation:

Step1: Use point - slope form

The point - slope form of a line is $y - y_1=m(x - x_1)$, where $(x_1,y_1)$ is a point on the line and $m$ is the slope. Given $m = \frac{2}{3}$ and the point $(5,-6)$, we substitute into the formula:
$y-(-6)=\frac{2}{3}(x - 5)$
Simplify the left - hand side: $y + 6=\frac{2}{3}(x - 5)$

Step2: Eliminate the fraction

Multiply both sides of the equation by 3 to get rid of the fraction:
$3(y + 6)=3\times\frac{2}{3}(x - 5)$
$3y+18 = 2(x - 5)$

Step3: Expand the right - hand side

Expand $2(x - 5)$: $3y+18=2x-10$

Step4: Rearrange into standard form

The standard form of a line is $Ax + By=C$, where $A$, $B$, and $C$ are integers and $A\geq0$.
Subtract $2x$ from both sides: $- 2x+3y+18=-10$
Subtract 18 from both sides: $-2x + 3y=-28$
Multiply both sides by - 1 to make $A>0$: $2x-3y = 28$

Answer:

The equation of the line in standard form is $2x - 3y=28$