While the exact formula of the line will always allow you to determine the area, you don't actually NEED that, so it's best not to assume.
Statement 1, the slope of p is -5/3. A line whose only restriction is its slope can cross the y-axis anywhere. If it's through the origin (y-intercept = 0), the area of the region bounded by p, the x-axis, and the y-axis is 0 since the three lines meet at a single point and a point has no area. The area will increase the further away from the origin (0,0) you move p - the bigger or smaller the value of the y-intercept, for example y-intercept = 100 or y-intercept = -100. On your scratch paper, or in your head, draw several lines with the same negative slope crossing the y-axis at different points, and it's apparent that the bounded area will differ for each point crossed. Since a slope of -5/3 doesn't provide a unique solution to the question in the stem, Statement 1 is not sufficient, so eliminate choices A and D.
Statement 2, the y-intercept of p is 10. A line whose only restriction is its y-intercept can have any slope. If the slope is 0, it is horizontal and the area bounded by p, the x-axis, and the y-axis is infinite because there is no closed shape. Now, imagine or draw other lines p with different slopes, intersecting the y-axis at point (0,10). The steeper the slope of p (the greater the absolute value of the slope), the smaller the bounded area; and the gentler the slope of p (the smaller the absolute value of the slope), the greater the bounded area. Since a y-intercept of 10 doesn't provide a unique solution to the question in the stem, Statement 2 is not sufficient, so eliminate choice B.
Combining the statements, if a line has a slope of -5/3 and crosses the y-axis at (0,10), then the bounded region has only one possible shape and thus a unique area. Choice C is the correct answer.
The strategy to use here is to take each statement and to "draw and redraw" - on scratch paper or in your head. This is the same as plugging in for plane geometry problems. In evaluating statement 1, you are keeping the slope constant and "plugging in" different y-intercepts to see what happens. In evaluating statement 2, you are keeping the y-intercept constant and "plugging in" different slopes to see what happens.
Terry Serres
The Princeton Review
Certified Instructor, Content Developer
Online Tutoring
[email protected]
