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.

Would you like me to explain this further or provide more details?

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 $x = 0$!