Question
A line passed through the point (0,1) and has a posititve slope. Which of these points could that line NOT pass through. Select the TWO that apply.
A. (1, 15)
B. (5, -2)
C. (12, 3)
D. (-3, 1)
E. (-2, -5)
A. (1, 15)
B. (5, -2)
C. (12, 3)
D. (-3, 1)
E. (-2, -5)
Answers
GPT-4o mini
To determine which points the line with a positive slope that passes through the point (0, 1) cannot pass through, we consider the general form of the line's equation. Since the line passes through (0, 1), we can express the equation in slope-intercept form as:
\[ y = mx + 1 \]
where \( m \) is the slope of the line and is positive (\( m > 0 \)).
For a point \((x, y)\) to lie on this line, it must satisfy the equation:
\[ y = mx + 1 \]
This means that as \( x \) increases, \( y \) also increases, since \( m \) is positive.
Let's analyze each option one by one:
A. (1, 15)
- If \( x = 1 \), then \( y = m(1) + 1 = m + 1 \). Since \( m > 0 \), \( m + 1 > 1 \), so \( y \) could be 15 if \( m \) is sufficiently large. This point could be on the line.
B. (5, -2)
- If \( x = 5 \), then \( y = m(5) + 1 = 5m + 1 \). Since \( m > 0 \), \( 5m + 1 > 1 \), which means \( y \) can never be -2 because -2 is less than 1. Therefore, this point could NOT be on the line.
C. (12, 3)
- If \( x = 12 \), then \( y = m(12) + 1 = 12m + 1 \). Since \( m > 0 \), \( 12m + 1 > 1 \), so \( y \) can be 3 if \( m \) is small enough. This point could be on the line.
D. (-3, 1)
- If \( x = -3 \), then \( y = m(-3) + 1 = -3m + 1 \). This implies \( y \) could be equal to 1, specifically when \( m = 0 \). However, since we are given that \( m > 0 \), \( -3m + 1 < 1 \). Therefore, the line can be at y = 1 but never actually reaches this point while still having a positive slope. Thus, this point could NOT be on the line.
E. (-2, -5)
- If \( x = -2 \), then \( y = m(-2) + 1 = -2m + 1 \). Since \( m > 0 \), \( -2m + 1 < 1 \). Given that -5 is much less than 1, this point can be on the line if \( m \) is sufficiently large. Therefore, it is possible for this point to be on the line.
Thus, the two points that the line with a positive slope passing through (0, 1) cannot pass through are:
**B. (5, -2)** and **D. (-3, 1)**.
\[ y = mx + 1 \]
where \( m \) is the slope of the line and is positive (\( m > 0 \)).
For a point \((x, y)\) to lie on this line, it must satisfy the equation:
\[ y = mx + 1 \]
This means that as \( x \) increases, \( y \) also increases, since \( m \) is positive.
Let's analyze each option one by one:
A. (1, 15)
- If \( x = 1 \), then \( y = m(1) + 1 = m + 1 \). Since \( m > 0 \), \( m + 1 > 1 \), so \( y \) could be 15 if \( m \) is sufficiently large. This point could be on the line.
B. (5, -2)
- If \( x = 5 \), then \( y = m(5) + 1 = 5m + 1 \). Since \( m > 0 \), \( 5m + 1 > 1 \), which means \( y \) can never be -2 because -2 is less than 1. Therefore, this point could NOT be on the line.
C. (12, 3)
- If \( x = 12 \), then \( y = m(12) + 1 = 12m + 1 \). Since \( m > 0 \), \( 12m + 1 > 1 \), so \( y \) can be 3 if \( m \) is small enough. This point could be on the line.
D. (-3, 1)
- If \( x = -3 \), then \( y = m(-3) + 1 = -3m + 1 \). This implies \( y \) could be equal to 1, specifically when \( m = 0 \). However, since we are given that \( m > 0 \), \( -3m + 1 < 1 \). Therefore, the line can be at y = 1 but never actually reaches this point while still having a positive slope. Thus, this point could NOT be on the line.
E. (-2, -5)
- If \( x = -2 \), then \( y = m(-2) + 1 = -2m + 1 \). Since \( m > 0 \), \( -2m + 1 < 1 \). Given that -5 is much less than 1, this point can be on the line if \( m \) is sufficiently large. Therefore, it is possible for this point to be on the line.
Thus, the two points that the line with a positive slope passing through (0, 1) cannot pass through are:
**B. (5, -2)** and **D. (-3, 1)**.