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y x = 3 2 x O 2 y = 2 2 4 x x O y = 2 2 2 4 x x
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2; (b) that are 3 units from
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2; see Fig. 12-23(a). 3 and x 3; see Fig. 12-23(b).
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4; see Fig. 12-23(c).
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y x=3 4 2 O (3, 0) 4 2 x (5, 0) y x=4
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Fig. 12-23
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12.17 Applying principles 3 and 4 Graph and describe the locus whose equation is (a) y
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Solutions
(a) The locus is a line whose y-intercept is 1 and whose slope equals 3. See Fig. 12-24(a). (b) The locus is a line which passes through the origin and has slope 3. See Fig. 12-24(b). 2 (c) The locus is a line which passes through the point (1, 1) and has slope 3. See Fig. 12-24(c). 4
y 4 2 (0, 1) x O y y
1 3x
1; (b) y
3 2 x;
3 . 4
x = 3 2
1 x 3
4 2 x O
4 2 (1, 1) x 4 x O
1 = 4 y 1 x
y = 1 4 x
y = 3 x = 2
y = 3 x
x = 4 2 4
y (a)
y (b)
y (c)
Fig. 12-24
CHAPTER 12 Analytic Geometry
12.18 Applying principle 5 Graph and give the equation of the locus of points (a) 2 units from the origin; (b) 2 units from the locus of x2 y2 9.
Solutions
(a) The locus is a circle whose equation is x2 y2 4. See Fig. 12-25(a).
(b) The locus is a pair of circles, each 2 units from the circle with center at O and radius 3. Their equations are x2 y2 25 and x2 y2 1. See Fig. 12-25(b).
y 4 2 x O x2 + y2 = 4 x 2 4 x O 2 4 4 2 x y
y (a)
y (b)
Fig. 12-25
12.6 Areas in Analytic Geometry
12.6A Area of a Triangle
If one side of a triangle is parallel to either coordinate axis, the length of that side and the length of the altitude to that side can be found readily. Then the formula A 1bh can be used. 2 If no side of a triangle is parallel to either axis, then either 1. The triangle can be enclosed in a rectangle whose sides are parallel to the axes (Fig. 12-26), or 2. Trapezoids whose bases are parallel to the y-axis can be formed by dropping perpendiculars to the x-axis (Fig. 12-27).
Fig. 12-26
Fig. 12-27
The area of the triangle can then be found from the areas of the figures so formed: 1. In Fig. 12-26, area (^ABC) 2. In Fig. 12-27, area (^ABC) area(rectangle ADEF) [area(^ ABD) area (^ BCE) area (^ACF)]. area(trapezoid ABED) area(trapezoid BEFC) area (trapezoid DFCA).
12.6B Area of a Quadrilateral
The trapezoid method described above can be extended to finding the area of a quadrilateral if its vertices are given.
CHAPTER 12 Analytic Geometry
SOLVED PROBLEMS
12.19 Area of a triangle having a side parallel to an axis Find the area of the triangle whose vertices are A(1, 2), B(7, 2), and C(5, 4).
Solution
From the graph of the triangle (Fig. 12-28), we see that b A 1bh 1(6)(2) 6. 2 2 7 1 6 and h 4 2 2. Then
4 2 x O y 2 A(1, 2)
C(5, 4) h b 4 B(7, 2) x
Fig. 12-28
12.20 Area of a triangle having no side parallel to an axis Find the area of ^ABC whose vertices are A(2, 4), B(5, 8) and C(8, 2) (a) using the rectangle method; (b) using the trapezoid method.
Solutions
See Fig. 12-29.
B(5, 8)
B(5, 8)
4 2 x O y
A (2, 4) D 2 4 (a) C(8, 2) x
4 2 x O y
A (2, 4) C(8, 2) G 2 4 (b) H J x
Fig. 12-29
(a) Area of rectangle DEFC Area of ^DAC Area of ^ABE Area of ^BCF So area(^ABC)
1 2 bh 1 2 bh 1 2 bh
1 2 (2)(6) 1 2 (3)(4) 1 2 (3)(6)
6(6) 6. 6. 9.
36. Then:
area(DEFC)
area(^DAC
^ABE
^BCF)
CHAPTER 12 Analytic Geometry
1 2 h(b 1 2 (3)(2 1 2 (6)(2 1 2 (3)(4
(b) Area of trapezoid ABHG Area of trapezoid BCJH Area of trapezoid ACJG Then area(^ABC)
br) 8) 4)
15. 18. area(ACJG) 18 15 18 15.
area(ABHG)
area(BCJH)
12.7 Proving Theorems with Analytic Geometry
Many theorems of plane geometry can be proved with analytic geometry. The procedure for proving a theorem has two major steps, as follows: 1. Place each figure in a convenient position on a graph. For a triangle, rectangle, or parallelogram, place one vertex at the origin and one side of the figure on the x-axis. Indicate the coordinates of each vertex (Fig. 12-30).
Triangle y C(c, h) h x c A (0, 0) y b x B(b, 0) x A (0, 0) y b B(b, 0) y D (0, h) h x x c A (0, 0) y Rectangle y C(b, h) D(c, h) h x b B(b, 0) C(b + c, h) Parallelogram
Fig. 12-30
2. Apply the principles of analytic geometry. For example, prove that lines are parallel by showing that their slopes are equal; or that lines are perpendicular by showing that their slopes are negative reciprocals of each other. Use the midpoint formula when the midpoint of a segment is involved, and use the distance formula to obtain the lengths of segments.
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