left of P such that PS
= PRo By drawing lines throughS andR perpendicular
to the x-axis and a line through P parallel to the x-axis, we form two right triangles L..PQR and L..PTS.By Theorem 5.19 we can prove L..PQR
:::=L..STP. Therefore, QR = TP and PQ = ST
and
QR - TP PQ - ST
ButQR ST PQ
PQ=ml and m2=-TP=-QR Therefore
m2-
1
ml
Thus, we have shown that the slopes of two nonvertical perpendicular lines are negative reciprocals of each other. The converse of this fact can be proved by retracing our steps.
Theorem 13.5. Two nonvertical lines il and i2 are perpendicular if and only if their slopes are negative reeiprfJCals ofeachuther:
il
1. i2 ~ ml=--
1 m2 ExercisesIn Exs. I -6 find the slope of the line passing through the points.
1. (7,3),(4,-3). 2. (-3,3),(3,-4).
3. (4,1), (-1, 6). 4. (-3,0), (I, 2).
5. (- 3, 2), (2, 1).
6. (6, - 4), (2, - 3).
In Exs. 7-12 check to see if AB
"
CD or if AB
1.CD.
7. A(-4,2),B(4,-I),C(3,2),D(II,-I).
R. A(2, - 2), B(- 3,4), C(5, 0), D(- I, - 5).
9. A (2,4), B(O, 0), C(- 2, 1), D(- 1, 3).
10. A(-1,-2),B(-2,-4),C(3,-2),D(I,-I).
II. A (0,4), B(O, 11), C(3,
- 3), D(4, - 3).
12. A (- 4, - 5), B(- 3,4), C(3, I), D(- 6,2).
---
--- ---
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Ex.n. Ex. 18 -
~ Ii:
The lines joining the midpoints of the opposite sides of a quadrilateral
','~
bisect each other.
The diagonals of a rhombus intersect at right angles.
.
376
FUNDAMENTALS OF COLLEGE GEOMETRYIn Ex. 13-15 prove the points are collinear.
13. A (3,4), B(4, 6), C(2, 2).
14. A(-3, O),B(-I, I), C(3, 3).
IS. A(-4,-6),B(0,-7),C(-8,-5).
Prove the following theorems analytically (figures are drawn as an aid for your proofs):
16. The segment joining the midpoints of two sides of a triangle is parallel to the third side.
17. The median of a trapezoid is parallel to the bases and has a measure equal to half the sum of the measure of the bases.
18. The diagonals of a square intersect at right angles.
.Y
\
y
D(d,c) C(b, c) D (0, b) C (b, b)
A (0, 0) B (b, 0)
A(O, 0) B (a, 0) x
19.
20.
y
H(a,bJ
.Y
C(a + b, c)
x A (0, 0)
Ex. 19. Ex. 20.
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COORDINATE GEOMETRY
377
3.10. The graph of a condition. The graph of a condition imposed on two variables is the set of points whose coordinates (x, y) satisfy the condition.
Often the condition is given in equation or inequation form. To graph an equation (or inequation) or condition in x and y means to draw its graph. To obtain it, we draw the figure which represents all the points whose coordinates satisfy the condition. The graph might be lines, rays, segments, triangles, circles, half-planes, or subsets of each. There follow examples which illustrate the relationship between a condition and its graph.
Example. Draw the graphs of the point P (x, y) which satisfy the following conditions: (a) x
=
3; (b) x = 0; (c) y = -5; (d) y ~ 0; (e) I < x ~ 4; (j) OP ~ 2.Solution:
.Y .Y
x 0 x
0
Example (a). Example (b).
.Y
x 0
ExamPle (c).
378
FUNDAMENTALS OF COLLEGE GEOMETRY Y-1 a -1
y
a
y
~
---
x
Example (d).
x
Fxrrmple (e).
x
Example ({).
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COORDINATE GEOMETRY 379 Exercises
Draw and describe the graphs which satisfy the following conditions.
1. x ~ O.
3. x > 0 and y > O.
5. x > 0 and y < O.
7. x>Oorx<-2.
9. -1 < x < 3 and 2 < y < 4.
11. x = 1 and y ~ O.
13. iYl < 2.
15. x> 2andy <-1.
2. y < O.
4. x < 0 and y > O.
6. -1 < x < 2.
8. Y < 1 or y > 4.
10. 2 ~ x ~ 4 and 3 ~ y ~ 5.
12. Ixl = 3.
14. Iyl > 2.
16. x> O,y > O,andy=x.
13.11. Equation of a line. The equation of a line in a plane is an equation in two variables, such as x and y, which is satisfied by every point on the line and is not satisfied by any point not on the line. The form of the equation will depend upon the data used in determining the line. A straight line is determined geometrically in several ways. If two points are used to deter- mine the line, the equation of the line will have a different form than if one point and a direction were used. We will consider some of the more common forms of the equation for a straight line.
13.12. Horizontal and vertical lines. If a line II is parallel to the y-axis, then every point on II has the same x-coordinate (see Fig. 13.13). If this
x-coordinate is a, then the point P (x, y) is on II if and only if x =
a.In like manner. y = b is the equation of 1"2'a line through (0, b) parallel to the x-axis.
Fig. 13.13.
13.13. Point-slope form of equation of a line. One of the simplest ways in which a line is determined is to know the coordinates of a point through which it passes and the slope of the line.
380 FUNDAMENTALS OF COLLEGE GEOMETRY
Consider a nonvertical line I passing through PI (Xb YI) with a slope m (see Fig. 13.14). Let P(x, y) be any point other than Plan the given line. p'
y
x
Fig. 13.14.
will lie on I ifand only
if the slope of WI is m; that is P (x, y) is on I
~ slope of WI is m, orP(x,y)isonl~~=m y-y X-XI and
P(X, y) is on I ~ Y-YI = m(x-xI)'
Theorem 13.6. For each point PI (Xb YI) and for each number m, the equation of the line through PI with slope m is Y - YI = m (x - XI)' This equation is called the point-slope form of an equation of a line.
ExamPle. Find the equation of the line which contains the point with co- ordinates (2, -3) and has a slope of5.
Solution:
Y-YI = m(x-xI) Y- (-3) = 5(x-2)
y + 3 = 5x - 10 5x-y-I3 = 0
Example.
Find the slope of the line whose equation is 5x - 2y = 11..~~?
I,
Solution:
follows:
COORDINATE GEOMETRY 381 We must first reduce our equation to the point-slope form, as
5x-2y = 11
5x-11 = 2y (Why?) 2y=5x-II (Why?)
=5(x-V)
y = t(x - V) (Why?) or
y-O=t(x-li)
Comparing this equation with the standard point-slope form equation, we find that the line passes through (¥, 0) and has a slope on.
13.14. Two-point form of equation of a line. The equation of a straight line that passes through two points can be obtained by use of the point-slope form and the equation for the slope of a line through two points. Thus, if PI (Xb 1"1) and P2 (X2' Y2) are coordinates of two points through which the line passes, the slope of the line is m= (YI-Y2)/(XI-X2) and substituting this value for m in the point-slope form, we get the equation
YI-Y2 Y-YI =-(X-XI) XI-X2
This is called the tu'o-point form of the equation of a straight line.
Exercises
Find an equation for each of the lines described.
1. The line contains the point with coordinates (7, 3) and its slope is 4.
2. The line contains the point with coordinates (-2, -5) and has a slope of3.
3. The line has a slope of -2 and passes through (-6, 8).
4. The line has an inclination of 45 and passes through (3, 5).
5. The line passes through (-9, -3) and is parallel to the x-axis.
6. The line contains the point (5, -7) and is perpendicular to the x-axis.
7. The line contains the points with coordinates (4, 7) and (6,11).
8. The line contains the point whose coordinates are (-1, 1) and has an inclination of 90.
In Exs. 9-16 find the slope of each of the lines with the following equations.
9. 3x-y = 7.
11. 5x + 3y = 9.
10. 2x+y12. Y = x. = 8...
382
FUNDAMENTALS OF COLLEGE GEOMETRY13.2x=y. 14. y=2x-7.
15. y = -3. 16. x = 6.
In Exs. 17-22, points A (-2,4), B (2, -4), C(6, 6) are vertices of ,6,ABe.
17. Find the equation of n.
18. Find the equation of BC.
19. Find the equation of the median drawn from C.
20. Find the equation of the median drawn from A.
21. Find the equation of the perpendicular bisector of AB.
22. Find the equation of the perpendicular bisector of Be.
y
X'
Y' Exs.17-22.
C(6,6)
x
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The x-intercept and y-intercept 13.15. Intercept form of equation of a line.
of a line are defined as the coordinates of the points where the line crosses the x-axis and the y-axis respectively. The terms are also used for the distances these points are from the origin. The context of the statement will make clear if a coordinate or distance is meant.
If the x-intercept and y-intercept of a line are respectively a and b (Fig. 13.15), the coordinates of the points of inter- section of the line and axes are (a,O) and (0, b). Using the two-point form,
y
---
0
Fig. 13.15.
x
:.1",.,..".
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...
or
y-b=-(x-O) b-O O-a y-b=_b -x
a
This may be reduced tobx
+ ay = ab
and by dividing both sides of the equation by ab, we get x Y
-+- = 1 .a b
which is the interceptform of the equation of a line.
ExamPle. Draw the graph of the line L whose equation is 3x- 4y-
12 = O.
Solution: 3x-4y-12 = O.
Adding 12 to both sides,
3x-4y = 12.
Dividing both sides by 12,
::_~ =
]4 ~
or
x Y