ANALYTIC GEOMETRY

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

xxyym

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-xy-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

xxyym

112

121 xx

xxyyyy

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-twothe as known is which

xxxxyyy-y

us give willform slope-point the to it ing substitutand

xxyym 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.

0xbym

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 thereforeand b-ymx us give

willxby sidesboth

gmultiplyin and 0-xb-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.

abm

0xabby

1by

ax

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

1by

ax us give willabby sides

both dividing and abaybxbecome willpositive terms the allmake to equation the arranging

-reby Then .bxab-ay us give willaby sidesboth

gmultiplyin and 0xab-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

sinpb

bpsin

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

sincosm or ,cot

tan1m

sinpx

sincosy

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 .

BA

AB

ABtan

22 BA

Bsin

22 BA

Acos

22 BA

0 BA

C yBA

B xBA

A222222

BA

C yBA

B xBA

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).

0344

34y3

34x5 ,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 173174

17y

174x

be wouldform, normal the in lines, required the of equations The174

17y

174x

1164

116y

1164x

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.

2211

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