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The problem in your image asks to find the equation of a line in slope-intercept form, given the slope \(-\frac{2}{3}\) and the point \((6, -10)\) through which the line passes.

### Step-by-Step Solution:

1. **Slope-Intercept Form**:
   The slope-intercept form of a line is:
   \[
   y = mx + b
   \]
   where \(m\) is the slope, and \(b\) is the y-intercept.

2. **Substitute the slope**:
   You're given the slope \(m = -\frac{2}{3}\). So the equation becomes:
   \[
   y = -\frac{2}{3}x + b
   \]

3. **Substitute the point (6, -10)**:
   Since the point \((6, -10)\) lies on the line, substitute \(x = 6\) and \(y = -10\) into the equation to solve for \(b\):
   \[
   -10 = -\frac{2}{3}(6) + b
   \]
   Simplifying:
   \[
   -10 = -4 + b
   \]

4. **Solve for \(b\)**:
   Add 4 to both sides to isolate \(b\):
   \[
   b = -10 + 4 = -6
   \]

5. **Write the final equation**:
   Now that we have \(b = -6\), substitute it into the slope-intercept form equation:
   \[
   y = -\frac{2}{3}x - 6
   \]

### Final Answer:
The equation of the line is:
\[
y = -\frac{2}{3}x - 6
\]

Would you like further details on any part of the solution or have any questions?

### 5 Related Questions:
1. How do you convert an equation from slope-intercept form to point-slope form?
2. Can you find the x-intercept of this line using the equation?
3. How would you graph the equation of this line?
4. What is the equation of a line perpendicular to this one that passes through the same point?
5. How can you derive the equation of a line given two points?

### Tip:
Always double-check the calculation of the y-intercept by substituting both the point and slope back into the slope-intercept form.

A line has a slope of -2/3 and passes through the point (6, -10). Write its equation in slope-intercept form.

Learn how to find the equation of a line with slope -2/3 passing through the point (6, -10) in slope-intercept form. This example covers key algebraic concepts such as the slope-intercept formula and solving for the y-intercept. Suitable for students in Grades 8-10.

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