SAT Mc Graw Hill 2011

CHAPTER 10 / ESSENTIAL GEOMETRY SKILLS 381

Answer Key 4: Coordinate Geometry

Concept Review 4

1. (−2, 1)


2. (0.5, 2)


3. 2/5


4. Your line should be vertical (straight up and
   down) and pass through B.


5. The x-coordinate of all the points is 3, so  1 can be
   described by the equation x=3.
   6. 2 should be horizontal (straight across) and pass
   through B.
   7. 3 should be a horizontal (straight across) line
   one unit below the x-axis.


6. (−2, 5)


7. (1, 8) (Notice that the new segment must be the
   same length as AB,but its slope is the negative
   reciprocal of AB’s slope, that is, −5/2.)


8. (8, 5)


9. The area of the rectangle is 108, and its length DC
   is 14 − 2 =12. So its height is 108/12 =9. There-
   fore, k− 1 =9, and k=10.
   m= 14
   n= 10
   p= 1

12.ACis the hypotenuse of a right triangle with legs

of 9 and 12. This is a 3-4-5 triangle times 3—a

9-12-15 triangle—so ACis 15. The perimeter of

the rectangle is 9 + 12 + 9 + 12 =42. So the ratio

of the diagonal to the perimeter is 15/42 =5/14.

13. The slope of DBis 3/4, or .75.
    14.ACand DBintersect at the midpoint of each seg-
    ment. The midpoint of ACis
    ((14 +2)/2, (10 +1)/2) =(8, 5.5).


14. (26, 19)


15. Use the left side of the
    triangle as the base.
    This way, the base is
    vertical and the height
    is horizontal, so the
    lengthsare easier to
    find. The base is 4
    units and the height
    is 7 units, so the area is (4 ×7)/2 =14.


16. The reflection of the triangle over the line x= 3
    is shown above. The “leftmost” point has an
    x-coordinate of 0.


17. No matter how the triangle is reflected, the area
    remains the same. The area is still 14.

x

y

(6, 7)

(-1, 3)

(-1, -1)

b = 4

x = 3

h = 7

SAT Practice 4

1. B The vertices of a
   square must always be
   listed in consecutive
   order, so point Cmust
   follow consecutively
   after Band can be in
   either position shown
   in the figure at right.
   Therefore, Ccan be at
   (0, 2) or (6, 2).


2. A The horizontal line passing through (1, 8) is
   the line y=8, and the vertical line passing through
   (−3, 4) is the line x=−3. So they intersect at (−3, 8).


3. E The distance from (0, 0) to (−3, 4) is 5, which
   is the radius of the circle. Therefore, any point that
   is 5 units from the origin is also on the circle. Each
   of the given points is also 5 units from the origin.


4. E The distance from (3, −7) to (8, 5) is

.

The circumference is 2πr= 26 π.

83 5 7 5 12 1 3

(^2222)
()− +−−()()=+=

5. D ∆ABChas a base of 4 and height of 8, so its area
   is (4 ×8)/2 =16. Since the base of ∆CDEis 2, its height
   must be 16 if it is to have the same area as ∆ABC.The
   y-coordinate of E,then, must be 16 or −16.


6. E Draw a rectangle
   around the quadrilateral
   as in the figure at right.
   The rectangle has an
   area of 9 × 8 =72. If we
   “remove” the areas of
   the four right triangles
   from the corners, the
   area of the shaded re-
   gion will remain.
   72 − 4 − 12 −12.5 −7.5 = 36
   7.^3 ⁄ 8 The y-coordinate
   of point A is 6,
   which means the
   “rise” from Oto Ais
   6. Since the slope of
    1 is^1 ⁄ 2 , the “run”
   must be 12.
   The “run” from Oto Bis 12 + 4 =16, and the “rise”
   is 6, so the slope of  2 is 6 /16 =3/8.

x

y

1

C 2

A

B C 1

x

y

(^4) 12.5
7.5
12
9
8
y
x
y = 6
O

1
2

AB 4

6

12