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Let's work through this step by step. The problem involves finding the x- and y-intercepts of the line \( 6x - 4y = -12 \), which gives points B (x-intercept) and C (y-intercept) on the coordinate plane, along with vertex A at \( (0, 0) \).

### Step 1: Find Point B (x-intercept)
To find the x-intercept, set \( y = 0 \) in the equation \( 6x - 4y = -12 \):

\[
6x - 4(0) = -12 \implies 6x = -12 \implies x = \frac{-12}{6} = -2
\]

So, Point B is \( (-2, 0) \).

### Step 2: Find Point C (y-intercept)
To find the y-intercept, set \( x = 0 \) in the equation \( 6x - 4y = -12 \):

\[
6(0) - 4y = -12 \implies -4y = -12 \implies y = \frac{-12}{-4} = 3
\]

So, Point C is \( (0, 3) \).

### Step 3: Sketch the Triangle
Now that we have the three points:
- Point A: \( (0, 0) \)
- Point B: \( (-2, 0) \)
- Point C: \( (0, 3) \)

These points form a right triangle with the x- and y-axes.

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Here are 5 related questions for practice:
1. How do you find the x- and y-intercepts of the line \( 3x + 5y = 15 \)?
2. Can you derive the equation of the line passing through points B and C?
3. How do you calculate the area of a triangle using its vertices' coordinates?
4. What is the distance between points B and C in this triangle?
5. How would the equation of the line change if the intercepts were different?

**Tip**: Always remember that the x-intercept occurs when \( y = 0 \) and the y-intercept occurs when \( x = 0 \)!

Point A is one vertex of triangle ABC. Point B is the x-intercept of 6x - 4y = -12, and point C is the y-intercept. What are points B and C? Sketch the triangle in the coordinate plane.

Learn how to find the x-intercept and y-intercept of the line 6x - 4y = -12. This problem includes calculating the coordinates of points B and C and sketching the triangle ABC on a coordinate plane. Perfect for students in Grades 8-10.

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