Question

If a line can be represented by the equation y+2=1/3(x−12), which of the following sets of points lies on the line? (1 point)
Responses

(12,−2) and (15,−1)
left parenthesis 12 comma negative 2 right parenthesis and left parenthesis 15 comma negative 1 right parenthesis

(−12,2) and (−9,3)
left parenthesis negative 12 comma 2 right parenthesis and left parenthesis negative 9 comma 3 right parenthesis

(12,−2) and (13,1)
left parenthesis 12 comma negative 2 right parenthesis and left parenthesis 13 comma 1 right parenthesis

(−12,2) and (−11,3)
left parenthesis negative 12 comma 2 right parenthesis and left parenthesis negative 11 comma 3 right parenthesis

Answer 1

To determine which set of points lies on the line represented by the equation \( y + 2 = \frac{1}{3}(x - 12) \), we can rewrite the equation in slope-intercept form \( y = mx + b \).

Starting with the original equation:

\[
y + 2 = \frac{1}{3}(x - 12)
\]

Distributing the right side:

\[
y + 2 = \frac{1}{3}x - 4
\]

Now, subtracting 2 from both sides:

\[
y = \frac{1}{3}x - 6
\]

Now we have our slope-intercept form: \( y = \frac{1}{3}x - 6 \).

Now we can test each set of points to see if they lie on this line.

1. **Check Point (12, -2)**:
\[
y = \frac{1}{3}(12) - 6 = 4 - 6 = -2 \quad \text{(True)}
\]
**Check Point (15, -1)**:
\[
y = \frac{1}{3}(15) - 6 = 5 - 6 = -1 \quad \text{(True)}
\]
Both points (12, -2) and (15, -1) lie on the line.

2. **Check Point (-12, 2)**:
\[
y = \frac{1}{3}(-12) - 6 = -4 - 6 = -10 \quad \text{(False)}
\]
**Check Point (-9, 3)**:
\[
y = \frac{1}{3}(-9) - 6 = -3 - 6 = -9 \quad \text{(False)}
\]
Neither point (-12, 2) nor (-9, 3) lies on the line.

3. **Check Point (12, -2)**:
\[
y = \frac{1}{3}(12) - 6 = 4 - 6 = -2 \quad \text{(True)}
\]
**Check Point (13, 1)**:
\[
y = \frac{1}{3}(13) - 6 \approx 4.33 - 6 \approx -1.67 \quad \text{(False)}
\]
Point (12, -2) lies on the line, but (13, 1) does not.

4. **Check Point (-12, 2)**:
\[
y = \frac{1}{3}(-12) - 6 = -4 - 6 = -10 \quad \text{(False)}
\]
**Check Point (-11, 3)**:
\[
y = \frac{1}{3}(-11) - 6 \approx -3.67 - 6 \approx -9.67 \quad \text{(False)}
\]
Neither point (-12, 2) nor (-11, 3) lies on the line.

**Conclusion**: The only set of points that lies on the line is:
**(12, -2) and (15, -1)**.