Question

what is the equation of the line that passes through the point (3, -8) and has a slope of -4/3?

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, \(x_1 = 3\), \(y_1=-8\) and \(m =-\frac{4}{3}\).
Substitute these values into the point - slope formula:
\(y-(-8)=-\frac{4}{3}(x - 3)\)
Simplify the left - hand side: \(y + 8=-\frac{4}{3}(x - 3)\)

Step2: Convert to slope - intercept form (optional, but to get a more standard form)

Distribute the \(-\frac{4}{3}\) on the right - hand side:
\(y+8=-\frac{4}{3}x+4\)
Subtract 8 from both sides of the equation:
\(y=-\frac{4}{3}x + 4-8\)
Simplify the right - hand side:
\(y=-\frac{4}{3}x-4\)

We can also leave it in the point - slope form or convert it to the standard form \(Ax + By=C\). Let's convert \(y=-\frac{4}{3}x-4\) to standard form.
Multiply through by 3 to get rid of the fraction:
\(3y=-4x-12\)
Add \(4x\) to both sides:
\(4x + 3y=-12\)

Answer:

The equation of the line in slope - intercept form is \(y =-\frac{4}{3}x-4\), in point - slope form is \(y + 8=-\frac{4}{3}(x - 3)\) and in standard form is \(4x+3y=-12\) (the most common forms are slope - intercept or standard form, and the slope - intercept form is \(y=-\frac{4}{3}x - 4\))