Question
Use the image to answer the question.
A coordinate plane shows both axes ranging from negative 6 to 6 in unit increments. A line with arrows on both the sides is plotted. The graph shows a line starting at left parenthesis 6 comma negative 1.5 right parenthesis and ending at left parenthesis negative 6 comma 1.5 right parenthesis.
The graph of a proportional linear equation is shown. Which of the following points is not contained on the graph?
(1 point)
Responses
(16,−4)
left parenthesis 16 comma negative 4 right parenthesis
(36,−9)
left parenthesis 36 comma negative 9 right parenthesis
(−10,40)
left parenthesis negative 10 comma 40 right parenthesis
(−24,6)
A coordinate plane shows both axes ranging from negative 6 to 6 in unit increments. A line with arrows on both the sides is plotted. The graph shows a line starting at left parenthesis 6 comma negative 1.5 right parenthesis and ending at left parenthesis negative 6 comma 1.5 right parenthesis.
The graph of a proportional linear equation is shown. Which of the following points is not contained on the graph?
(1 point)
Responses
(16,−4)
left parenthesis 16 comma negative 4 right parenthesis
(36,−9)
left parenthesis 36 comma negative 9 right parenthesis
(−10,40)
left parenthesis negative 10 comma 40 right parenthesis
(−24,6)
Answers
Answered by
GPT-4o mini
To determine which point is not contained on the graph of the linear equation defined by the line extending from (6, -1.5) to (-6, 1.5), we first need to establish the equation of the line.
1. **Find the slope (m) of the line**:
- From points (6, -1.5) and (-6, 1.5),
- Change in y: \(1.5 - (-1.5) = 1.5 + 1.5 = 3\)
- Change in x: \(-6 - 6 = -12\)
- Slope, \(m = \frac{change\ in\ y}{change\ in\ x} = \frac{3}{-12} = -\frac{1}{4}\)
2. **Use point-slope form to find the equation of the line**:
- Using point (6, -1.5):
- \(y - (-1.5) = -\frac{1}{4}(x - 6)\)
- Simplifying:
\[
y + 1.5 = -\frac{1}{4}x + \frac{6}{4}
\]
\[
y = -\frac{1}{4}x + \frac{3}{2} - 1.5
\]
\[
y = -\frac{1}{4}x + 0
\]
- Therefore, the equation is \(y = -\frac{1}{4}x\) which is in the form \(y = mx\).
3. **Substituting the points to check which is not on the line**:
- For each point, we will substitute \(x\) into the equation and see if it yields the corresponding \(y\).
a) **Point (16, −4)**:
\[
y = -\frac{1}{4}(16) = -4 \quad (\text{This point is on the line})
\]
b) **Point (36, −9)**:
\[
y = -\frac{1}{4}(36) = -9 \quad (\text{This point is on the line})
\]
c) **Point (−10, 40)**:
\[
y = -\frac{1}{4}(-10) = \frac{10}{4} = 2.5 \quad (40 \neq 2.5, \text{ so this point is NOT on the line})
\]
d) **Point (−24, 6)**:
\[
y = -\frac{1}{4}(-24) = 6 \quad (\text{This point is on the line})
\]
Thus, the point that is **not contained** on the graph is:
**(−10, 40)**.
1. **Find the slope (m) of the line**:
- From points (6, -1.5) and (-6, 1.5),
- Change in y: \(1.5 - (-1.5) = 1.5 + 1.5 = 3\)
- Change in x: \(-6 - 6 = -12\)
- Slope, \(m = \frac{change\ in\ y}{change\ in\ x} = \frac{3}{-12} = -\frac{1}{4}\)
2. **Use point-slope form to find the equation of the line**:
- Using point (6, -1.5):
- \(y - (-1.5) = -\frac{1}{4}(x - 6)\)
- Simplifying:
\[
y + 1.5 = -\frac{1}{4}x + \frac{6}{4}
\]
\[
y = -\frac{1}{4}x + \frac{3}{2} - 1.5
\]
\[
y = -\frac{1}{4}x + 0
\]
- Therefore, the equation is \(y = -\frac{1}{4}x\) which is in the form \(y = mx\).
3. **Substituting the points to check which is not on the line**:
- For each point, we will substitute \(x\) into the equation and see if it yields the corresponding \(y\).
a) **Point (16, −4)**:
\[
y = -\frac{1}{4}(16) = -4 \quad (\text{This point is on the line})
\]
b) **Point (36, −9)**:
\[
y = -\frac{1}{4}(36) = -9 \quad (\text{This point is on the line})
\]
c) **Point (−10, 40)**:
\[
y = -\frac{1}{4}(-10) = \frac{10}{4} = 2.5 \quad (40 \neq 2.5, \text{ so this point is NOT on the line})
\]
d) **Point (−24, 6)**:
\[
y = -\frac{1}{4}(-24) = 6 \quad (\text{This point is on the line})
\]
Thus, the point that is **not contained** on the graph is:
**(−10, 40)**.