Final answer:
The roads described by equations A (y = 3x - 10), B (x = 3), and C (y + 6 = 3(x - 15)) have slopes that are the negative reciprocal of the slope of Oak Street (6x + 18y = 5), making them perpendicular to it.
Step-by-step explanation:
The streets in MidTown are organized in a grid pattern, and we are looking for equations that represent roads perpendicular to Oak Street, which is described by the equation 6x + 18y = 5. To find roads that are perpendicular, we must identify equations with slopes that are the negative reciprocal of the slope of Oak Street. The slope of Oak Street is determined by rewriting it in slope-intercept form, y = mx + b, where 'm' is the slope. First, we solve 6x + 18y = 5 for y:
18y = -6x + 5
y = -⅓x + ⅛
Thus, the slope of Oak Street is -⅓ (or -1/3). A perpendicular street will have a slope of 3, which is the negative reciprocal of -1/3. Let's evaluate each option:
- A. y = 3x − 10: This equation's slope is 3, which makes it perpendicular to Oak Street.
- B. x = 3: This is the equation of a vertical line, which is indeed perpendicular to Oak Street.
- C. y + 6 = 3(x − 15): When simplified, this equation gives a slope of 3, making it perpendicular to Oak Street.
- D. 3x + 9y = 8: When this equation is rewritten in slope-intercept form, the slope is 1/3, which is not the negative reciprocal of -1/3, so it's not perpendicular.
- E. 2x − 3y = 5: This equation simplifies to a slope of 2/3, which is also not the negative reciprocal of -1/3, so it's not perpendicular.
Options A, B, and C represent roads that are perpendicular to Oak Street.