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12.3 Graphing a quadrilateral If the vertices of a rectangle have the coordinates A(3, 1), B( 5, 1), C ( 5, its perimeter and area.
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The base and height of the rectangle are 8 and 4 (see Fig. 12-4). Hence, the perimeter is 2b 2(4) 24, and the area is bh (8)(4) 32.
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6 4 2
2 C( 5, 3) 4 D(3, 3) +2 +4
x 3 B( 2 1 , 2) 2 O 3 D y 3
A(4 1 , 2) 2
Fig. 12-4
Fig. 12-5
CHAPTER 12 Analytic Geometry
12.4 Graphing a triangle 1 1 If the vertices of a triangle have the coordinates A A 42, 2 B , B A 22, 2 B and C(1, 5), find its area.
Solution
The length of the base is BA 7 (see Fig. 12-5). The height is CD 7. Then A
1 2 bh 1 2 (7)(7)
12.2 Midpoint of a Segment
The coordinates (xm, ym) of the midpoint M of the line segment joining P(x1, y1) to Q(x2, y2) are xm
1 2 (x1
1 2 (y1
In Fig. 12-6, segment ym is the median of trapezoid CPQD, whose bases are y1 and y2. Since the length of a median is one-half the sum of the bases, ym 1(y1 y2). Similarly, segment xm is the median of trapezoid 2 ABQP, whose bases are x1 and x2; hence, xm 1(x1 x2). 2
Fig. 12-6
SOLVED PROBLEMS
12.5 Applying the midpoint formula If M is the midpoint of PQ, find the coordinates of (a) M if the coordinates of P and Q are P(3, 4) and Q(5, 8); (b) Q if the coordinates of P and M are P(1, 5) and M(3, 4).
Solutions
(a) xm (b) xm
1 2 (x1 1 2 (x1
1 2 (3
1 2 (1
4; ym
1 2 (y1
y2) 5; ym
1 2 (4 1 2 (y1
1 2 (5
x2), so 3
x2) and x2
y2), so 4
y2) and y2
12.6 Determining if segments bisect each other The vertices of a quadrilateral are A(0, 0), B(0, 3), C(4, 3), and D(4, 0). (a) Show that ABCD is a rectangle. (b) Show that the midpoint of AC is also the midpoint of BD. (c) Do the diagonals bisect each other Why
CHAPTER 12 Analytic Geometry
Solutions
(a) From Fig. 12-7, AB CD 3 and BC AD 4; hence, ABCD is a parallelogram. Since /BAD is a right angle, ABCD is a rectangle.
Fig. 12-7
(b) The coordinates of the midpoint of AC are x
The coordinates of the midpoint of BD are x 1 Hence, (2, 12) is the midpoint of both AC and BD.
1 2 (0 1 2 (0
4) 4)
2, y 2, y
1 2 (0 1 2 (3
3) 0)
12. 12.
(c) Yes, since the midpoints of both diagonals are the same point.
12.3 Distance Between Two Points
PRINCIPLE 1:
The distance between two points having the same ordinate (or y-value) is the absolute value of the difference of their abscissas. (Hence, the distance between two points must be positive.)
6 3.
Thus, the distance between the point P(6, 1) and Q(9, 1) is 9
PRINCIPLE 2:
The distance between two points having the same abscissa (or x-value) is the absolute value of the difference of their ordinates.
1 3.
Thus, the distance between the points P(2, 1) and Q(2, 4) is 4
PRINCIPLE 3:
The distance d between the points P1 (x1, y1) and P2(x2, y2) is d 2(x2 x1)2 (y2 y1)2 or d 2( x)2 ( y)2
The difference x2 x1 is denoted by the symbol x; the difference y2 y1 is denoted by y. Delta ( ) is the fourth letter of the Greek alphabet, corresponding to our d. The difference x and y may be positive or negative.
SOLVED PROBLEMS
12.7 Providing and using the distance formula (a) Prove the distance formula (Principle 3) algebraically. (b) Use it to find the distance between A(2, 5) and B(6, 8).
Solutions
(a) See Fig. 12-8. By Principle 1, P1S angle P1SP2, or and x2 x1 x. By Principle 2, P2S (P1S)2 (x2 2(x2 (P2S)2 x1)2 x1)2 (y2 (y2 y1)2 y1)2 y2 y1 y. Also, in right tri-
(P1P2)2 d2 d
CHAPTER 12 Analytic Geometry
Fig. 12-8
(b) The distance from A(2, 5) to B(6, 8) is found as follows: (x, y) B(6, 8) S x2 A(2, 5) S x1 d 2 (x2 d2 (6 6, y2 8 2, y1 5 x1)2 (y2 y1)2 2)2 (8 5)2 42
25 and d
12.8 Finding the distance between two points Find the distance between the points (a) ( 3, 5) and (1, 5); (b) (3, (6, 8); ( 3, 2) and (9, 3).
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