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12.12 Slope and inclination of a line (a) Find the slope of the line through ( 2, 1) and (4, 3). 4x x 15. 4.
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(b) Find the slope of the line whose equation is 3y
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(c) Find the inclination of the line whose equation is y
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(a) By principle 1, m (b) We may rewrite 3y (c) Since y x y2 x2 4x y1 x1 3 4 ( 1) ( 2)
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4, we have m
1; thus, tan i
12.13 Slopes of parallel or perpendicular lines 4 4 4 4 4 4 2 Find the slope of CD if (a) AB y CD and the slope of AB is 3; (b) AB ' CD and the slope of 4 AB is 3. 4
Solutions
(a) By Principle 8, slope of CD (b) By Principle 10, slope of CD
slope of AB 1 slope of AB
2 3.
1 3/4
4 . 3
12.14 Applying principles 9 and 11 to triangles and quadrilaterals Complete each of the following statements: (a) In quadrilateral ABCD, if the slopes of AB, BC, CD, and DA are 2, the quadrilateral is a ____ . (b) In triangle LMP, if the slopes of LM and MP are 5 and
Solutions
(a) Since the slopes of the opposite sides are equal, ABCD is a parallelogram. In addition, the slopes of adjacent sides are negative reciprocals; hence, those sides are and ABCD is a rectangle. (b) Since the slopes of LM and MP are negative reciprocals, LM ' MP and the triangle is a right triangle.
2, 1, and 2
2, respectively,
1 5,
then LMP is a ____ triangle.
12.15 Applying principle 12 (a) AB has a slope of 2 and points A, B, and C are collinear. What are the slopes of AC and BC (b) Find y if G(1, 4), H(3, 2), and J(9, y) are collinear.
CHAPTER 12 Analytic Geometry
Solutions
(a) By Principle 12, AC and BC have a slope of 2. (b) By Principle 12, slope of GJ y 4.
slope of GH. Hence
y 4 4 , so that 1 8
1 and
12.5 Locus in Analytic Geometry
A locus of points is the set of points, and only those points, satisfying a given condition. In geometry, a line or curve (or set of lines or curves) on a graph is the locus of analytic points that satisfy the equation of the line or curve. Think of the locus as the path of a point moving according to a given condition or as the set of points satisfying a given condition.
PRINCIPLE 1:
The locus of points whose abscissa is a constant k is a line parallel to the y-axis; its equation is x k. (See Fig. 12-19.) The locus of points whose ordinate is a constant k is a line parallel to the x-axis; its equation is y k. (See Fig. 12-19.)
PRINCIPLE 2:
y y y=k x=k x O y y x x
= O y mx
Fig. 12-19
Fig. 12-20
PRINCIPLE 3:
The locus of points whose ordinate equals the product of a constant m and its abscissa is a straight line passing through the origin; its equation is y mx.
The constant m is the slope of the line. (See Fig. 12-20.)
PRINCIPLE 4:
The locus of points whose ordinate and abscissa are related by either of the equations y mx b or y x y1 x1 m
where m and b are constants, is a line (Fig. 12-21). y In the equation y mx b, m is the slope and b is the y-intercept. The equation x the line passes through the fixed point (x1, y1) and has a slope of m.
y1 x1
m tells us that
x O y
y P(x, y) r x O y x x
y y1 =m x x1 (x1, y1) x
Fig. 12-21
Fig. 12-22
CHAPTER 12 Analytic Geometry
PRINCIPLE 5:
The locus of points such that the sum of the squares of the coordinates is a constant is a circle whose center is the origin. x2 y2 r2 y2 r2.
The constant is the square of the radius, and the equation of the circle is (see Fig. 12-22). Note that for any point P(x, y) on the circle, x2
SOLVED PROBLEMS
12.16 Applying principles 1 and 2 Graph and give the equation of the locus of points (a) whose ordinate is the y-axis; (c) that are equidistant from the points (3, 0) and (5, 0).
Solutions
(a) From Principle 2, the equation is y (b) From Principle 1, the equation is x (c) The equation is x