Jump to a New ChapterIntroduction to the SAT IIIntroduction to SAT II Math ICStrategies for SAT II Math ICMath IC FundamentalsAlgebraPlane GeometrySolid GeometryCoordinate GeometryTrigonometryFunctionsStatisticsMiscellaneous MathPractice Tests Are Your Best Friends
 8.1 The Coordinate Plane 8.2 Lines and Distance 8.3 Lines 8.4 Graphing Linear Inequalities 8.5 Other Important Graphs and Equations

 8.6 Coordinate Space 8.7 Key Formulas 8.8 Review Questions 8.9 Explanations
Explanations

1.      B

If you understand how to answer this question, you should be able to answer any question dealing with line segments on the Math IC. The question asks you to find the distance between two midpoints, so you first need to find where those midpoints are. By definition, a midpoint is a point equidistant from both ends of a line segment. So the midpoint of AD is 152 = 7.5 units from either end. To find the midpoint of BC, you first need to find the length of BC. From the figure, you can see that BC = ADABCD. Since it is given that AB + CD = 25 AD,

The midpoint of BC is 92 = 4.5 units from B and C. But in order to measure the distance between the midpoints of BC and AD, you must situate the midpoint of BC with regard to AD. You can do this by finding out the midpoint of BC’s distance from point A.

Therefore, the midpoint of BC is 4.5 + 4.5 = 9 units from a, whereas the midpoint of AD is 7.5 units from a. The distance between these two midpoints is 9 – 7.5 = 1.5.

2.      E

The slope of y = 3x + 4 is 3. A line perpendicular to this line has a slope that is the opposite of the reciprocal of 3, or –13. If the x-intercept of the desired line is 6, then the line contains the point (6, 0), and we have enough information to put it in point-slope form: (y – 0) = –13(x – 6). This simplifies to y = –13 x + 2.

3.      A

The slope of the line pictured is –126 = –2. The y-intercept of the line pictured is –2. So the equation of the line pictured is y = –2x – 2. This leaves three choices. The line in the picture is dotted; this means that the inequality is either < or >, not ≤ or ≥. Now we have two choices left. Because the region to the right and above is shaded, > is the correct inequality, but just to make sure, plug in a point or two. In this case, try the origin, (0,0). It is in the shaded region, and must therefore satisfy the inequality. Which inequality does is satisfy? It satisfies y > –2x – 2, so this is the correct choice.

4.      D

In this equation, y is raised to the first power, and x is squared. This should lead you to believe that the graph of this equation is a parabola (in the equation of a circle, both variables are squared). Simplify the equation to arrive at the standard form:

The equation in the question is the standard form of a parabola. From it we can see that –3 < 0, so the parabola opens downward, and the vertex is (2, 6).

 Jump to a New ChapterIntroduction to the SAT IIIntroduction to SAT II Math ICStrategies for SAT II Math ICMath IC FundamentalsAlgebraPlane GeometrySolid GeometryCoordinate GeometryTrigonometryFunctionsStatisticsMiscellaneous MathPractice Tests Are Your Best Friends
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