Questions: A line passes through the point (6,-3) and has a slope of 4/3. Write an equation in slope-intercept form for this line.

Solution Steps

To find the equation of a line in slope-intercept form \( y = mx + b \) given a point \((x_1, y_1)\) and a slope \( m \), we can use the point-slope form of the equation \( y - y_1 = m(x - x_1) \) and then solve for \( y \) to convert it to slope-intercept form.

1. Start with the point-slope form of the equation: \( y - y_1 = m(x - x_1) \).
2. Substitute the given point \((6, -3)\) and the slope \( \frac{4}{3} \) into the equation.
3. Solve for \( y \) to convert the equation to slope-intercept form \( y = mx + b \).

We are given a point \((6, -3)\) and a slope \(m = \frac{4}{3}\).

We start with the point-slope form of the equation of a line: \[ y - y_1 = m(x - x_1) \] Substituting the given values: \[ y - (-3) = \frac{4}{3}(x - 6) \]

This simplifies to: \[ y + 3 = \frac{4}{3}(x - 6) \] Distributing the slope: \[ y + 3 = \frac{4}{3}x - 8 \]

Now, we isolate \(y\): \[ y = \frac{4}{3}x - 8 - 3 \] \[ y = \frac{4}{3}x - 11 \]

Final Answer