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 \(m=\frac{1}{3}\).
Substitute these values into the point - slope formula: \(y-(-5)=\frac{1}{3}(x - 6)\)

Step2: 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 (\(y=mx + b\)):
\(y=\frac{1}{3}x-2 - 5\)
\(y=\frac{1}{3}x-7\)
We can also write it in the standard form \(Ax+By = C\) (where \(A\), \(B\), and \(C\) are integers and \(A\geq0\)):
Multiply through by 3 to get rid of the fraction: \(3y=x - 21\)
Rearrange to get \(x-3y=21\) (or we can also leave it in slope - intercept form)

Answer:

The equation of the line is \(y=\frac{1}{3}x - 7\) (or \(x-3y = 21\))