A number line is a line on which distances from a point are marked off in equal units, positively in one direction and negatively in the other. The origin is the zero point from which distances are measured. Figure 12-1 shows a horizontal number line.
Fig. 12-1
The graph shown in Fig. 12-2 is formed by combining two number lines at right angles to each other so that their zero points coincide. The horizontal number line is called the x-axis, and the vertical number line is the y-axis. The point where the two lines cross each other is, again, called the origin.
Fig. 12-2
A point is located on a graph by its coordinates, which are its distances from the axes. The abscissa or x-coordinate of a point is its distance from the y-axis. The ordinate or y-coordinate of a point is its distance from the x-axis.
When the coordinates of a point are stated, the x-coordinate precedes the y-coordinate. Thus, the coordinates of point P in Fig. 12-2 are written (4, 3); those for Q are (–4,–3). Note the parentheses.
The quadrants of a graph are the four parts cut off by the axes. These are numbered I, II, III, and IV in a counterclockwise direction, as shown in Fig. 12-2.
12.1 Locating points on a graph
Give the coordinates of the following points in Fig. 12-3:
Fig. 12-3
(b) M
(c) O
(d) L
(e) N
(f) G
(g) P
(h) Q
(i) R
(j) S
Solutions
(a) (2, 2)
(b) (–3, 3)
(c) (0, 0)
(d) (–4,–3)
(e) 
(f) 
(g) (0, 3)
(h) (–2, 0)
(i) (0,–1
)
(j) (2, 0)
12.2 Coordinates of points in the four quadrants
What are the signs of the coordinates of (a) a point in quadrant I; (b) a point in quadrant II; (c) a point in quadrant III; (d) a point in quadrant IV? Show which coordinate has a sign and which zero value for a point between quadrants (e) IV and I; (f) I and II; (g) II and III; (h) III and IV.
Solutions
(a) (+, +)
(b) (–, +)
(c) (–,–)
(d) (+,–)
(e) (+, 0)
(f) (0, +)
(g) (–, 0)
(h) (0,–)
12.3 Graphing a quadrilateral
If the vertices of a rectangle have the coordinates A(3, 1), B(–5, 1), C (–5,–3), and D (3–3), find its perimeter and area.
Solution
The base and height of the rectangle are 8 and 4 (see Fig. 12-4). Hence, the perimeter is 2b + 2h = 2(8) + 2(4) = 24, and the area is bh = (8)(4) = 32.
Fig. 12-4
If the vertices of a triangle have the coordinates A (4,–2), B(–2,–2) 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 = bh = (7)(7) = 24.
Fig. 12-5
The coordinates (xm, ym) of the midpoint M of the line segment joining P(x 1, y 1)m to Q(x 2, y 2) are
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, 
Similarly, segment xm is the median of trapezoid ABQP, whose bases are x1 and x2; hence, 
Fig. 12-6
12.5 Applying the midpoint formula
If M is the midpoint of 
, 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 = (x1 + x2) = (3 + 5) = 4; ym = (y1 + y2) = (4 + 8) = 6.
(b) xm = (x1 + x2), so 3 = (1 + x2) and x2 = 5; ym = (y1 + y2), so 4 = (5 + y2) and y2 = 3.
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 
is also the midpoint of 
.
(c) Do the diagonals bisect each other? Why?
(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 are x = (0 + 4) = 2, y = (0 + 3) = 1. The coordinates of the midpoint of are x = (0 + 4) = 2, y = (3 + 0) = 1. Hence, (2, 1) is the midpoint of both and .
(c) Yes, since the midpoints of both diagonals are the same point.
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.)
Thus, the distance between the point P(6, 1) and Q(9, 1) is 9–6 = 3.
PRINCIPLE 2: The distance between two points having the same abscissa (or x-value) is the absolute value of the difference of their ordinates.
Thus, the distance between the points P(2, 1) and Q(2, 4) is 4–1 = 3.
PRINCIPLE 3: The distance d between the points P (x1, y1) and P2(x2, y2) is
The difference x2–x1 is denoted by the symbol Δx; the difference y2–y1 is denoted by Δy. Delta (Δ) is the fourth letter of the Greek alphabet, corresponding to our d. The difference Δx and Δy may be positive or negative.
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, P 1 S = x2–x1 = Δx. By Principle 2, P 2 S = y2–y1 = Δy. Also, in right triangle P1SP2,
(b) The distance from A(2, 5) to B(6, 8) is found as follows:
12.8 Finding the distance between two points
Find the distance between the points (a) (–3, 5) and (1, 5); (b) (3,–2) and (3, 4); (c) (3, 4) and (6, 8); (–3, 2) and (9,–3).
Solutions
(a) Since both points have the same ordinate (or y-value), d = x2–x1 = 1–(–3) = 4
(b) Since both points have the same abscissa (or x-value), d = y 2–y 1 = 4–(–2) = 6
(c) 
(d) 
12.9 Applying the distance formula to a triangle
(a) Find the lengths of the sides of a triangle whose vertices are A(1, 1), B(1, 4), and C(5, 1).
(b) Show that the triangle whose vertices are G(2, 10), H(3, 2), and J(6, 4) is a right triangle.
Solutions
See Fig. 12-9.
Fig. 12-9
(a) AC = 5–1 = 4 and AB = 4–1 = 3; 
(b) (GJ)2 = (6–2)2 + (4–10)2 = 52; (HJ)2 = (6–3)2 + (4–2)2 = 13; (GH)2 = (2–3)2 + (10–2)2= 65. Since (GJ)2 + (HJ)2 = (GH)2, ΔGHJ is a right triangle.
12.10 Applying the distance formula to a parallelogram
The coordinates of the vertices of a quadrilateral are A(2, 2), B(3, 5), C(6, 7), and D(5, 4). Show that ABCD is a parallelogram.
Solution
See Fig. 12-10, where we have
Thus, 
≐ 
and 
≐ 
. Since opposite sides are congruent, ABCD is a parallelogram.
Fig. 12-10
12.11 Applying the distance formula to a circle
A circle is tangent to the x-axis and has its center at (6, 4). Where is the point (9, 7) with respect to the circle?
Solution
Since the circle is tangent to the x-axis, 
in Fig. 12-11 is a radius. By Principle 2, AQ = 4.
Fig. 12-11
By Principle 3, 
Since 
is greater than 4, 
is greater than a radius so B is outside the circle.
PRINCIPLE 1: If a line passes through the points P 1 (x 1 y 1) and P2 (x2, y2), then
PRINCIPLE 2: The line whose equation is y = mx + b has slope m.
PRINCIPLE 3: The slope of a line equals the tangent of its inclination.
The inclination i of a line is the angle above the x-axis that is included between the line and the positive direction of the x-axis (see Fig. 12-12). In the figure,
Fig. 12-12
The slope is independent of the order in which the end points are selected. Thus,
PRINCIPLE 4: If a line slants upward from left to right, its inclination i is an acute angle and its slope is positive (Fig. 12-13).
Fig. 12-13
PRINCIPLE 5: If a line slants downward from left to right, its inclination is an obtuse angle and its slope is negative (Fig. 12-14).
Fig. 12-14
PRINCIPLE 6: If a line is parallel to the x-axis, its inclination is 0° and its slope is 0 (Fig. 12-15).
Fig. 12-15
PRINCIPLE 7: If a line is perpendicular to the x-axis, its inclination is 90° and it has no slope (Fig. 12-16).
Fig. 12-16
PRINCIPLE 8: Parallel lines have the same slope.
In Fig. 12-17, l║l’; hence, corresponding angles i and i’ are equal, and tan i = tan i’ or m = m’, where m and m’ are the slopes of 1 and 1’.
Fig. 12-17
PRINCIPLE 9: Lines having the same slope are parallel to each other. (This is the converse of Principle 8.)
PRINCIPLE 10: Perpendicular lines have slopes that are negative reciprocals of each other. (Negative reciprocals are numbers, such as 
and–
, whose product is–1.)
Thus in Fig. 12-18, if l 
l’, then m =–1/m’ or mm’ =–1, where m and m’ are the slopes of l and’.
Fig. 12-18
PRINCIPLE 11: Lines whose slopes are negative reciprocals of each other are perpendicular. (This is the converse of Principle 10.)
Collinear points are points which lie on the same straight line. Thus, A, B, and C are collinear points here:
PRINCIPLE 12: The slope of a straight line is constant all along the line.
Thus if 
above is a straight line, the slope of the segment from A to B equals the slope of the segment from C to Q.
PRINCIPLE 13: If the slope of a segment between a first point and a second equals the slope of the segment between either point and a third, then the points are collinear.
12.12 Slope and inclination of a line
(a) Find the slope of the line through (–2,–1) and (4, 3).
(b) Find the slope of the line whose equation is 3y–4x = 15.
(c) Find the inclination of the line whose equation is y = x + 4.
Solutions
(a) By principle 1, 
(b) We may rewrite 3y–4x = 15 as y = 
x + 5, from which m =
(c) Since y = x + 4, we have m = 1; thus, tan i = 1 and i = 45°.
12.13 Slopes of parallel or perpendicular lines
Find the slope of 
if (a) 
|| and the slope of is 
; (b) and the slope of is 
.
Solutions
(a) By Principle 8, slope of = slope of = .
(b) By Principle 10, slope of 
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 , 
, , and 
are –2, and–2, respectively, the quadrilateral is a 
.
(b) In triangle LMP, if the slopes of 
and 
are 5 and–
, then LMP is a triangle.
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 and are negative reciprocals, 
and the triangle is a right triangle.
12.15 Applying principle 12
(a) has a slope of 2 and points A, B, and C are collinear. What are the slopes of and ?
(b) Find y if G(1, 4), H(3, 2), and J(9, y) are collinear.
(a) By Principle 12, and have a slope of 2.
(b) By Principle 12, slope of 
= slope of Hence 
so that 
and y =–4.
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.)
PRINCIPLE 2: The locus of points whose ordinate is a constant k is a line parallel to the x-axis; its equation isy = k. (See Fig. 12-19.)
Fig. 12-19
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.)
Fig. 12-20
PRINCIPLE 4: The locus of points whose ordinate and abscissa are related by either of the equations
where m and b are constants, is a line (Fig. 12-21).
Fig. 12-21
In the equation y = mx + b, m is the slope and b is the y-intercept. The equation 
tells us that the line passes through the fixed point (x1, y1) and has a slope of m.
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.
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 + y2 = r2.
Fig. 12-22
12.16 Applying principles 1 and 2
Graph and give the equation of the locus of points (a) whose ordinate is–2; (b) that are 3 units from the y-axis; (c) that are equidistant from the points (3, 0) and (5, 0).
Solutions
(a) From Principle 2, the equation is y =–2; see Fig. 12-23(a).
(b) From Principle 1, the equation is x = 3 and x =–3; see Fig. 12-23(b).
(c) The equation is x = 4; see Fig. 12-23(c).
Fig. 12-23
12.17 Applying principles 3 and 4
Graph and describe the locus whose equation is (a) 
+ 1; (b) 
(c) 
Solutions
(a) The locus is a line whose y-intercept is 1 and whose slope equals 
. See Fig. 12-24(a).
(b) The locus is a line which passes through the origin and has slope 
. See Fig. 12-24(b).
(c) The locus is a line which passes through the point (1, 1) and has slope . See Fig. 12-24(c).
Fig. 12-24
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).
Fig. 12-25
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 = bh can be used.
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
Fig. 12-26
2. Trapezoids whose bases are parallel to the y-axis can be formed by dropping perpendiculars to the x-axis (Fig. 12-27).
Fig. 12-27
The area of the triangle can then be found from the areas of the figures so formed:
1. In Fig. 12-26, area (ΔABC) = area(rectangle ADEF)–[area(Δ ABD) + area (ΔBCE) + area (ΔACF)].
2. In Fig. 12-27, area (ΔABC) = area(trapezoid ABED) + area(trapezoid BEFC)– area (trapezoid DFCA).
The trapezoid method described above can be extended to finding the area of a quadrilateral if its vertices are given.
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 = 7–1 = 6 and h = 4–2 = 2. Then A = bh = (6)(2) = 6.
Fig. 12-28
12.20 Area of a triangle having no side parallel to an axis
Find the area of ΔAABC 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.
Fig. 12-29
(a) Area of rectangle DEFC = bh = 6(6) = 36. Then:
Area of ΔDAC = bh = (2)(6) = 6.
Area of ΔABE = bh = (3)(4) = 6.
Area of ΔBCF = bh = (3)(6) = 9.
So area(ΔABC) = area(DEFC)–area(ΔDAC + ΔABE + ΔBCF) = 36–(6 + 6 + 9) = 15.
(b) Area of trapezoid ABHG = h(b + B’) = (3)(4 + 8) = 18.
Area of trapezoid BCJH = (3)(2 + 8) = 15.
Area of trapezoid ACJG = (6)(2 + 4) = 18.
Then area(ΔABC) = area(ABHG) + area(BCJH)–area(ACJG) = 18 + 15–18=15.
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).
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.
12.21 Proving a theorem with analytic geometry
Using analytic geometry, prove that the diagonals of a parallelogram bisect each other.
Solution
Given: 
ABCD, diagonals and .
To Prove: and bisect each other.
Plan: Use the midpoint formula to obtain the
coordinates of the midpoints of the diagonals
Place ABCD with vertex A at the origin and side along the x-axis (Fig. 12-31). Then the vertices have the coordinates A(0, 0), B(a, b), C(a + c, b), and D(c, 0).
Fig. 12-31
By the midpoint formula, the midpoint of has the coordinates 
and the midpoint of has the coordinates 
Then the diagonals bisect each other, since the midpoints of both diagonals are the same point.
12.1. State the coordinates of each lettered point in Fig. 12-32.
(12.1)
Fig. 12-32
12.2. Plot each of the following points:
A(–2,–3)
B(–3, 2
C(0,–1)
D(–3, 0)
E(3,-4)
F 
G(0, 3)
H 
(12.2)
12.3. Plot the following points: A(2, 3), B(–3, 3), C(–3,–2), D(2,–2). Then find the perimeter and area of square ABCD.
12.4. Plot the following points: A(4, 3), B(–1, 3), C(–3,–3), D(2,–3). Then find the area of parallelogram ABCD and triangle BCD
(12.3, 12.4)
12.5. Find the midpoint of the segment joining
(a) (0, 0) and (8, 6)
(b) (0, 0) and (5, 7)
(c) (0, 0) and (–8, 12)
(d) (14, 10) and (0, 0)
(e) (–20,–5) and (0, 0)
(f) (0, 4) and (0, 16)
(g) (8, 0) and (0,–2)
(h) (–10, 0) and (0,–5)
(i) (3, 4) and (7, 6)
(j) (–2,–8) and (–4,–12)
(k) (7, 9) and (3, 3)
(l) (2,–1) and (–2,–5)
(12.5)
12.6. Find the midpoints of the sides of a triangle whose vertices are
(a) (0, 0), (8, 0), (0, 6)
(b) (–6, 0), (0, 0), (0, 10)
(c) (12, 0), (0,–4), (0, 0)
(d) (3, 5), (5, 7), (3, 11)
(e) (4, 0), (0,–6), (–4, 10)
(f) (–1,–2), (0, 2), (1,–1)
(12.5)
12.7. Find the midpoints of the sides of the quadrilateral whose successive vertices are
(a) (0, 0), (0, 4), (2, 10), (6, 0)
(b) (–3, 5), (–1, 9), (7, 3), (5,–1)
(c) (–2, 0), (0, 4), (6, 2), (0,–10)
(d) (–3,–7), (–1, 5), (9, 0), (5,–8)
(12.5)
12.8. Find the midpoints of the diagonals of the quadrilateral whose successive vertices are
(a) (0, 0), (0, 5), (4, 12), (8, 1)
(b) (–4,–1), (–2, 3), (6, 1), (2,–8)
(c) (0,–5), (0, 1), (4, 9), (4, 3)
(12.5)
12.9. Find the center of a circle if the end points of a diameter are
(a) (0, 0) and (–4, 6)
(b) (–1, 0) and (–5,–12)
(c) (–3, 1) and (0,–5)
(d) (0, 0) and (2a, 2b)
(e) (a, b) and (3a, 5 b)
(f) (a, 2b) and (a, 2c)
(12.5)
12.10. If M is the midpoint of , find the coordinates of
(a) M if the coordinates of A and B are A(2, 5) and B(6, 11)
(b) A if the coordinates of M and B are M(1, 3) and B(3, 6)
(c) B if the coordinates of A and M are A(–2, 1) and M(2,–1)
(12.5)
12.11. The trisection points of are B and C. Find the coordinates of
(a) B if the coordinates of A and C are A(1, 2) and C(3, 5)
(b) D if the coordinates of B and C are B(0, 5) and C(1f, 4)
(c) A if the coordinates of B and C are B(0, 6) and C(2, 3)
(12.5)
12.12. A(0, 0), B(0, 5), C(6, 5), and D(6, 0) are the vertices of quadrilateral ABCD.
(a) Prove that ABCD is a rectangle.
(b) Show that the midpoints of and have the same coordinates.
(c) Do the diagonals bisect each other? Why?
(12.6)
12.13. The vertices of ΔABC are A(0, 0), B(0, 4), and C(6, 0).
(a) If is the median to find the coordinates of D and the midpoint of
(b) If 
is the median to find the coordinates of E and the midpoint of
(c) Do the medians, and , bisect each other? Why?
(12.6)
12.14. Find the distance between each of the following pairs of points:
(a) (0, 0) and (0, 5)
(b) (4, 0) and (–2, 0)
(c) (0,–3) and (0, 7)
(d) (–6,–1) and (–6, 11)
(e) (5, 3) and (5, 8.4)
(f) (–1.5, 7) and (6, 7)
(g) 
(h) (a, b) and (2a, b)
(12.8)
12.15. Find the distances separating pairs of the following collinear points:
(a) (5,–2), (5, 1), (5, 4)
(b) (0,–6), (0,–2), (0, 12)
(c) (–4, 2), (–3, 2), (0, 2)
(d) (0, b), (a, b), (3a, b)
(12.8)
12.16. Find the distance between each of the following pairs of points:
(a) (0, 0) and (5, 12)
(b) (–3,–4) and (0, 0)
(c) (0,–6) and (9, 6)
(d) (4, 1) and (7, 5)
(e) (–3,–6), and (3, 2)
(f) (2, 3) and (–10, 12)
(g) (2, 2) and (5, 5)
(h) (0, 5) and (–5, 0)
(i) (3, 4) and (4, 7)
(j) (–1,–1) and (1, 3)
(k) (–3, 0) and 
(l) (a, 0) and (0, a)
(12.8)
12.17. Show that the triangles having the following vertices are isosceles triangles:
(a) A(3, 5), B(6, 9), and C(2, 6)
(b) D(2, 0), E(6, 0), and F(4, 4)
(C) G(5,–5), H(–2,–2), and J(8, 2)
(d) K(7, 0), L(3, 4), and M(2,–1)
(12.9)
12.18. Which of the triangles having the following vertices are right triangles?
(a) A(7, 0), B(6, 3), and C(12, 5)
(b) D(2, 0), E(5, 2), and F(1, 8)
(c) G(1,–1), H(5, 0), and J(3, 8)
(d) K(–4, 0), L(–2, 4), and M(4,–1)
(12.9)
12.19. The vertices of ΔABC are A(–2, 2), B(4, 4), and C(8, 2). Find the length of the median to (a) ; (b) ; (c) .
(12.9)
12.20. (a) The vertices of quadrilateral ABCD are A(0, 0), B(3, 2), C(7, 7), and D(4, 5). Show that ABCD is a parallelogram.
(12.10)
(b) The vertices of quadrilateral DEFG are D(3, 5), E(1, 1), F(5, 3), and G(7, 7). Show that DEFG is a rhombus.
(c) The vertices of quadrilateral HJKL are H(0, 0), J(4, 4), K(0, 8), and L(–4, 4). Show that HJKL is a square.
12.21. Find the radius of a circle that has its center at
(a) (0, 0) and passes through (–6, 8)
(b) (0, 0) and passes through (3,–4)
(c) (0, 0) and passes through (–5, 5)
(d) (2, 0) and passes through (7,–12)
(e) (4, 3) and is tangent to the y-axis
(f) (–1, 7) and is tangent to the line x =–4
(12.11)
12.22. A circle has its center at the origin and a radius of 10. State whether each of the following points is on, inside, or outside of this circle: (a) (6, 8); (b) (–6, 8); (c) (0, 11); (d) (–10, 0); (e) (7, 7); (f) (–9, 4); (g) (9, 
).
(12.11)
12.23. Find the slope of the line through each of the following pairs of points:
(a) (0, 0) and (5, 9)
(b) (0, 0) and (9, 5)
(c) (0, 0) and (6, 15)
(d) (2, 3) and (6, 15)
(e) (–2,–3) and (7, 15)
(f) (–2,–3) and (2, 1)
(g) (3,–4) and (5, 6)
(h) (0, 0) and (–4, 8)
(i) (3,–9) and (0, 0)
(j) (0,–2) and (8, 10)
(k) (–1,–5) and (1,–7)
(l) (–3,–4) and (–1,–2)
(12.12)
12.24. Find the slope of the line whose equation is
(a) y = 3x–4
(b) y = 4x–3
(c) y =– x + 5
(d) y = 8–7x
(e) y = 5x
(f) y = 5
(g) 2y = 6x–10
(h) 2y = 10x–6
(i) 3y =–12x + 6
(j) 3y = 12–2x
(k) y + x = 21
(l) 2x = 12–y
(m) y = x–3
(n) y = 2x–6
(o) 
y = 7–x
(p) 
y + 2x = 1
(12.12)
12.25. Find the inclination, to the nearest degree, of each of the following lines:
(a) y = 3x–1
(b) y = x–1
(c) 2y = 5x + 10
(d) y = x + 5
(e) 5y = 5x–3
(f) y =–3
(12.12)
12.26. Find the slope of a line whose inclination is (a) 5°; (b) 17°; (c) 20°; (d) 35°; (e) 45°; (f) 73°; (g) 85°.
(12.12)
12.27. Find the inclination, to the nearest degree, of a line whose slope is (a) 0; (b) 0.4663; (c) 1, (d) 1.4281; (e) 
; (f) 
; (g) ; (h)1 (i) 
.
(12.12)
12.28. In hexagon ABCDEF of Fig. 12-33, . Which sides or diagonals have (a) positive slope; (b) negative slope; (c) zero slope; (d) no slope?
Fig. 12-33
12.29. Find the slope of a line that is parallel to a line whose slope is (a) 0; (b) has no slope; (c) 5; (d)–5; (e) 0.5; (f)–0.0005.
(12.13)
12.30. Find the slope of a line that is parallel to the line whose equation is
(a) y = 0
(b) x = 0
(c) x = 7
(d) y = 7
(e) y = 5x–2
(f) x + y = 5
(g) 3y–6x = 12
(12.13)
12.31. Find the slope of a line that is parallel to a line which passes through (a) (0, 0) and (2, 3); (b) (2,–1) and (5, 6); (c) (3, 4) and (5, 2); (d) (1, 2) and (0,–4).
(12.13)
12.32. Find the slope of a line that is perpendicular to a line whose slope is
(a)
(b) 1
(c) 3
(d)
(e) 0.1
(f)–1
(g) 
(h) 
(i) 0
(j) has no slope
(12.13)
12.33. Find the slope of a line that is perpendicular to a line which passes through (a) (0, 0) and (0, 5); (b) (0, 0) and (2, 1); (c) (0, 0) and (3,–1); (d) (1, 1) and (3, 3).
(12.13)
12.34. In rectangle DEFG, the slope of 
is . What is the slope of (a) 
; (b) 
; (c) 
?
(12.14)
12.35. In ABCD the slope of is 1 and the slope of is–. What is the slope of (a) ; (b) ; (c) the altitude of ; (d) the altitude to ?
(12.14)
12.36. The vertices of ΔABC are A(0, 5), B(3, 7), and C(5,–1). What is the slope of the altitude to (a) ; (b) ; (c) ?
(12.14)
12.37. Which of the following sets of points are collinear: (a) (2, 1), (4, 4), (8, 10); (b) (–1, 1), (2, 4), (6, 8); (c) (1,–1), (3, 4), (5, 8)?
(12.15)
12.38. What values of k will make the following trios of points collinear (a) A(0, 1), B(2, 7), C(6, k); (b) D(–1, 5), E(3, k), F(5, 11); (c) G(0, k), H(1, 1), I(3,–1)?
(12.15)
12.39. State the equation of the line or pair of lines which is the locus of points
(a) Whose abscissa is–5
(b) Whose ordinate is 3
(c) 3 units from the x-axis
(d) 5 units below the x-axis
(e) 4 units from the y-axis
(f) 3 units from the line x = 2
(g) 6 units above the line y =–2
(h) 1 unit to the right of the y-axis
(i) Equidistant from the lines x = 5 and x = 13
(12.18)
12.40. State the equation of the locus of the center of a circle that
(a) Is tangent to the x-axis at (6, 0)
(b) Is tangent to the y-axis at (0, 5)
(c) Is tangent to the lines x = 4 and x = 8
(d) Passes through the origin and (10, 0)
(e) Passes through (3, 7) and (9, 7)
(f) Passes through (3,–2) and (3, 8)
(12.18)
12.41. State the equation of the line or pair of lines which is the locus of points
(a) Whose coordinates are equal
(b) Whose ordinate is 5 more than the abscissa
(c) Whose abscissa is 4 less than the ordinate
(d) Whose ordinate exceeds the abscissa by 10
(e) The sum of whose coordinates is 12
(f) The difference of whose coordinates is 2
(g) Equidistant from the x–axis and y–axis
(h) Equidistant from x + y = 3 and x + y = 7
(12.16)
12.42. Describe the locus of each of the following equations:
(a) y = 2x + 5
(b) 
(c) 
(d) 
(e) x + y = 7
(f) 3 y = x
(12.17)
12.43. State the equation of a line which passes through the origin and has a slope of (a) 4; (b)–2; (c) ; (d)–; (e) 0.
(12.17)
12.44. State the equation of a line which has a y- intercept of
(a) 5 and a slope of 4
(b) 2 and a slope of–3
(c)–1 and a slope of
(d) 8 and is parallel to y = 3x–2
(e)–3 and is parallel to y = 7–4 x
(f) 0 and is parallel to y–2x = 8
(12.17)
12.45. State the equation of a line which has a slope of 2 and passes through (a) (1, 4); (b) (–2, 3); (c) (–4, 0); (d) (0,–7).
(12.17)
12.46. State the equation of a line
(a) Which passes through the origin and has a slope of 4
(b) Which passes through (0, 3) and has a slope of
(c) Which passes through (1, 2) and has a slope of 3
(d) Which passes through (–1,–2) and has a slope of
(e) Which passes through the origin and is parallel to a line that has a slope of 2
(12.18)
12.47. (a) Describe the locus of the equation x2 + y2 = 49.
(b) State the equation of the locus of points 4 units from the origin.
(c) State the equations of the locus of points 3 units from the locus of x2 + y2 = 25.
(12.18)
12.48. State the equation of the locus of point 5 units from (a) the origin; (b) the circle x2 + y2 = 16; (c) the circle x2 + y2 = 49.
(12.18)
12.49. What is the radius of the circle whose equation is (a) x2 + y2 = 9; (b) x2 + y2 = y
; (c) 9x2 + 9y2 = 36; (d) x2 + y2 = 3?
(12.18)
12.50. What is the equation of a circle whose center is the origin and whose radius is (a) 4; (b) 11; (c) ; (d) 1; (e) 
; (f) 
?
(12.18)
12.51. Find the area of ΔABC, whose vertices are A(0, 0) and
(a) B(0, 5) and C(4, 5)
(b) B(0, 5) and C(4, 2)
(c) B(0, 8) and C(–5, 8)
(d) B(0, 8) and C(–5, 12)
(e) B(6, 2) and C(7, 0)
(f) B(6,–5) and C (10, 0)
(12.19)
12.52. Find the area of a
(a) Triangle whose vertices are A(0, 0), B(3, 4), and C(8, 0)
(b) Triangle whose vertices are D(1, 1), E(5, 6), and F(1, 7)
(c) Rectangle three of whose vertices are H(2, 2), J(2, 6), and K(7, 2)
(d) Parallelogram three of whose vertices are L(3, 1), M(9, 1), and P(5, 5)
(12.19)
12.53. Find the area of ΔDEF, whose vertices are D(0, 0) and (a) E(6, 4) and F(8, 2); (b) E(3, 2) and F(6,–4); (c) E(–2, 3) and F(10, 7).
(12.20)
12.54. Find the area of a triangle whose vertices are (a) (0, 0), (2, 3), and (4, 1); (b) (1, 1), (7, 3), and (3, 6); (c) (–1, 2), (0,–2), and (3, 1).
(12.20)
12.55. The vertices of ΔABC are A(2, 1), B(8, 9), and C(5, 7). (a) Find the area of ΔABC. (b) Find the length of . (c) Find the length of the altitude to .
(12.20)
12.56. Find the area of a quadrilateral whose vertices are (a) (3, 3), (10, 4), (8, 7), and (5, 6); (b) (0, 4), (5, 8), (10, 6), and (14, 0); (c) (0, 1), (2, 4), (8, 10), and (12, 2).
12.57. Find the area of the quadrilateral formed by the lines
(12.19)
(a) x = 0, x = 5, y = 0, and y = 6
(b) x = 0, x = 7, y =–2, and y = 5
(c) x =–3, x = 5, y = 3, and y =–8
(d) x = 0, x = 6, y = 0, and y = x + 1
(e) y = 0, y = 4, y = x, and y = x + 4
(f) y = 0, y = 6, y = 2x, and y = 2x + 6
12.58. Prove each of the following with analytic geometry:
(12.21)
(a) A line segment joining the midpoints of two sides of a triangle is parallel to the third side.
(b) The diagonals of a rhombus are perpendicular to each other.
(c) The median to the base of an isosceles triangle is perpendicular to the base.
(d) The length of a line segment joining the midpoints of two sides of a triangle equals one-half the length of the third side.
(e) If the midpoints of the sides of a rectangle taken in succession are joined, the quadrilateral formed is a rhombus.
(f) In a right triangle, the length of the median to the hypotenuse is one-half the length of the hypotenuse.
