Math14 lesson 5
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Transcript of Math14 lesson 5
ANALYTIC GEOMETRY(Lesson 6)
Math 14 Plane and Analytic Geometry
STRAIGHT LINES/ FIRST DEGREE EQUATIONS
OBJECTIVES:At the end of the lesson, the student is expected to be
able to:• Define and determine the general equation of a line• Define and determine the different standard
equations of line• Determine the directed distance from a point to a
line• Determine the distance between parallel lines
STRAIGHT LINESA straight line is a locus of a point that
moves in a plane with constant slope. It may also be referred to simply as a line which contains at least two distinct points.
LINES PARALLEL TO A COORDINATE AXIS
If a straight line is parallel to the y-axis, its equation is x = k, where k is the directed distance of the line from the y-axis. Similarly, if a line is parallel to the x-axis, its equation is y = k, where k is the directed distance of the line from the x-axis.
DIFFERENT STANDARD FORMS OF THE EQUATION OF A STRAIGHT LINE
A. POINT-SLOPE FORM:If the line passes through the point (x1, y1), then the slope of the line is . Rewriting the equation we have which is the standard equation of the point-slope form.
1
1
xx
yym
11 xxmyy
The equation of the line through a given point P1 (x1, y1) whose slope is m.
y
x
111 y,xP
y,xP
m
form. slope
-point the as known is
which x-xmy-y or
y-yx-xm us give
willx-xby sidesboth
gmultiplyin and x-x
y-ym
formula, slopegsinU
11
11
1
1
1
EXAMPLE:Find the general equation of the line:a.through (2,-7) with slope of 2/5b.through the point (-3, 4) with slope of -2/5
B. TWO-POINT FORM:If the line passes through the points (x1, y1) and
(x2, y2), then the slope of the line is .
Substituting it in the point-slope formula, we have which the standard equation ofthe two-point form.
12
12
xx
yym
112
121 xx
xx
yyyy
The equation of the line through points P1 (x1, y1) and P2 (x2, y2)
y
x
111 y,xP
222 y,xP
m
y,xP
form. point-two
the as known is which
xxxx
yyy-y
us give willform slope
-point the to it ing substitutand
xx
yym slopethe ingUs
112
121
12
12
C. SLOPE-INTERCEPT FORM:Consider a line not parallel to either axes of the
coordinate axes. Let the slope of the line be m and intersecting the y-axis at point (0, b), then the slope of the line is . Rewriting the equation, we have
which is the standard equation of the slope-intercept form.
0x
bym
bmxy
EXAMPLE:Find the general equation of the line: a.passing through (4,-5) and (-6, 3)b.passing through (2,-3) and (-4, 5)
y
x
b,0
y,xP
m
form. intercept-slope
the as known is which
bmxy therefore
and b-ymx us give
willxby sidesboth
gmultiplyin and 0-x
b-ym
formula, slopegsinU
The equation of the line having the slope, m, and y-intercept (0, b)
EXAMPLE:a.Find the general equation of the line with slope 3 and y-intercept of 2/3. b.Express the equation 3x-4y+8=0 to the slope-intercept form and draw the line.
D. INTERCEPT FORM:Let the intercepts of the line be the points (a, 0)and (0, b). Then the slope of the line and itsequation is . Simplifying the equationwe have which is the standard equation ofthe intercept form.
a
bm
0xa
bby
1b
y
a
x
The equation of the line whose x and y intercepts are (a, 0) and (0, b) respectively.
y
x
b,0
0,a
m y,xP
form.
intercept the as known is which
1b
y
a
x us give willabby sides
both dividing and abaybx
become willpositive terms the all
make to equation the arranging
-reby Then .bxab-ay
us give willaby sidesboth
gmultiplyin and 0xa
b-b-y
formula, slope-point gsinU
EXAMPLE:Find the general equation of the line: a.with x-intercept of 2 and y-intercept of -3/4 b.through (-2, 7) with intercepts numerically equal but of opposite sign
E. NORMAL FORM:Suppose a line L, whose equation is to be found, has its distance from the origin to be equal to p. Let the angle of inclination of p be
o
b
y
x
p
L
sin
pb
b
psin
Since p is perpendicular to L, the slope of p is equal to the negative reciprocal of the slope L.
Substituting in the slope-intercept form,y = mx + b , we obtain
Simplifying, we have the normal form of the straight line
sin
cosm or ,cot
tan
1m
sin
px
sin
cosy
p y sin cos x
Reduction of the General Form to the Normal FormThe slope of the line Ax+By+C=0 is . The slope of p which is perpendicular to the line is therefore . Thus, .From Trigonometry, we obtain the values and . If we divide through the generalequation of the straight line by , we have
Transposing the constant to the right, we obtain This is of the normal form . Comparing the two equations, we note that .
B
A
A
B
A
Btan
22 BA
Bsin
22 BA
Acos
22 BA
0 BA
C y
BA
B x
BA
A222222
BA
C y
BA
B x
BA
A222222
22 BA
Cp
p y sin cos x
EXAMPLE: 1.Reduce 5x+3y-4=0 to the normal form.
2. Find the equation of a line parallel to the line 4x-y+8=0 passing at a distance ±3 from the point (-2,-4).
034
4
34
y3
34
x5 ,thus
34925BA
-4C 3B 5A
:Solution
22
4yx4424yx4
is 4,2 through gsinpas
and line given the to parallel line a of equation The
:Solution
1734yx4
or 17317
4
17
y
17
4x
be wouldform, normal the in lines, required the of equations The17
4
17
y
17
4x116
4
116
y
116
4x
have weform, normal the to ducingRe
PARALLEL AND PERPENDICULAR LINES The lines Ax+By+C=0 and Ax+By+K=0 are parallel
lines. But, the lines Ax+By+C=0 and Bx-Ay+K=0 are perpendicular lines.
EXAMPLE: Find the general equation of the line:a. through (-3, 8) parallel to the line 6x-5y+15=0 b. through (6,-1) and perpendicular to the line 4x-5y-6=0c. passing through (-1, 5) and parallel to the line through
(1 ,3) and (1,-4)
DIRECTED DISTANCE FROM A POINT TO A LINE The directed distance from the point P(x1, y1) to the
line Ax+By+C=0 is , where the sign of B isTaken into consideration for the sign of the . If
B>0, then it is and B<0, then it is . But if B=0, take the sign of A.
22
11
BA
CByAxd
22 BA 22 BA 22 BA
y
x
111 y,xP
222 y,xP
0CByAx 11
0d1
0d2
line the below is
point the 0,d if
line the above is
point the 0,d if
:note
EXAMPLE: a.Find the distance of the point (6,-3) from the line 2x-y+4=0.b.Find the equation of the bisector of the acute angle for the pair of lines L1: 11x+2y-7=0 and L2: x+2y+2=0.c.Find the distance between the lines 3x+y-12=0 and 3x+y-4=0
EXERCISES:1. Determine the equation of the line passing through (2, -3) and parallel to the line passing through (4,1) and (-2,2).2. Find the equation of the line passing through point (-2,3) and perpendicular to the line 2x – 3y + 6 = 03. Find the equation of the line, which is the perpendicular bisector of the segment connecting points (-1,-2) and (7,4).4. Find the equation of the line whose slope is 4 and passing through the point of intersection of lines x + 6y – 4 = 0 and 3x – 4y + 2 = 0
5. The points A(0, 0), B(6, 0) and C(4, 4) are vertices of triangles. Find:a. the equations of the medians and their intersection pointb. the equations of the altitude and their intersection pointc. the equation of the perpendicular bisectors of the sides and their intersection points
Exercises:1. Find the distance from the line 5x = 2y + 6 to the pointsa. (3, -5)b. (-4, 1)c. (9, 10)2. Find the equation of the bisector of the pair of acute angles formed by the lines 4x + 2y = 9 and 2x – y = 8.3. Find the equation of the bisector of the acute angles and also the bisector of the obtuse angles formed by the lines x + 2y – 3 = 0 and 2x + y – 4 = 0.
REFERENCES
Analytic Geometry, 6th Edition, by Douglas F. RiddleAnalytic Geometry, 7th Edition, by Gordon Fuller/Dalton Tarwater
Analytic Geometry, by Quirino and Mijares