Question

question
what is the equation of the line that passes through the point (6, -5) and has a slope of \\(\frac{1}{3}\\)?

Explanation:

Step1: Recall point - slope form

The point - slope form of a line is given by $y - y_1=m(x - x_1)$, where $(x_1,y_1)$ is a point on the line and $m$ is the slope of the line.
Here, $x_1 = 6$, $y_1=-5$ and $m=\frac{1}{3}$.

Step2: Substitute values into point - slope form

Substitute the values into the formula $y - y_1=m(x - x_1)$:
$y-(-5)=\frac{1}{3}(x - 6)$

Step3: Simplify the equation

Simplify the left - hand side: $y + 5=\frac{1}{3}(x - 6)$
Distribute the $\frac{1}{3}$ on the right - hand side: $y+5=\frac{1}{3}x-2$
Subtract 5 from both sides to get the slope - intercept form: $y=\frac{1}{3}x-2 - 5$
$y=\frac{1}{3}x-7$

We can also write it in standard form ($Ax+By = C$) by multiplying through by 3 to get rid of the fraction:
$3y=x - 21$
$x-3y=21$

Answer:

The equation of the line in slope - intercept form is $y=\frac{1}{3}x - 7$ (or in standard form $x-3y = 21$)