Lesson 6 straight line

STRAIGHT LINES/ FIRST DEGREE
EQUATIONS

Prepared by:
Prof. Teresita P. Liwanag – Zapanta
B.S.C.E., M.S.C.M., M.Ed. (Math-units), PhD-TM (on-going)

STRAIGHT LINES
A 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.

The equation of the line through a given
point P1 (x1, y1) whose slope is m.

y

x

Generally,

Considering points P(x, y) and P1 (x1, y1),

Therefore, m (x – x1) = y – y1

The equation of the line having the slope,
m, and y-intercept (0, b)

y P (x, y)

(0, b)

b

x

Generally,

Considering points P(x, y) and (0, b),

mx = y –b
Therefore, y = mx + b

The equation of the line whose x and y
intercepts are (a, 0) and (0, b) respectively.
y

B (0, b)

b-y P (x, y)
b

y
A (a, 0)
x
x a-x
a

IV. TWO POINT FORM

If the line passes through the points ( x1 , y1 ) and ( x 2 , y 2 ) , then the slope
y − y1
of the line is m = 2 . Substituting it in the point-slope formula, we have
x 2 − x1
y − y1
y − y1 = 2 ( x − x1 ) which is the standard equation of the two-point form.
x 2 − x1

The equation of the line through points
P1 (x1, y1) and P2 (x2, y2)

P2(x2, y2)
y

P (x, y)

P1(x1 , y1 )
x

Considering points P1 (x1, y1) and P2
(x2, y2),

1

Considering points P(x, y) and P1 (x1, y1),
2

Equation 1 = Equation 2

Examples:
I.Find the general equation of the line:
a. through (2, -7) with slope 2/5
b. with slope 3 and y-intercept 2/3
c. passing through (4, -5) and (-6, 3)
d. with x-intercept of 4 and y-intercept of -6
e. with slope 1/3 and passing through (5, -3)

f. passing through (-2, -7) and has its intercepts
numerically equal but of opposite signs

g. Determine the equation of the line passing
through (2, -3) and parallel to the line passing
through (4,1) and (-2,2).
h. Find the equation of the line passing through
point (-2,3) and perpendicular to the line
2x – 3y + 6 = 0
i. Find the equation of the line, which is the
perpendicular bisector of the segment connecting
points (-1,-2) and (7,4).
j. 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

II. 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 point
b. the equations of the altitude and their
intersection point
c. the equation of the perpendicular bisectors of
the sides and their intersection points

NORMAL FORM OF THE STRAIGTH LINE

A N
y

C (x1, y1)
P
y1

x
B
x1

Let: AB – given line
ON – line perpendicular to AB
C– point of intersection with coordinates
(x1,y1)
Recall: m = tanθ where: m – slope of line
θ – Inclination of line
mON = tanω therefore, mAB = -1/ tanω
mAB = - cotω
mAB = - cosω/sinω
x1 = Pcosω
y1 = Psinω

DISTANCE FROM A POINT TO LINE

y

P1 (x1,y1)
d

P

ω x

L L1

Sign Conventions:

a. The denominator is given by the sign of B.
b. The distance (d) is positive (+) if the point
P1 (x1 ,y1) is above the line.
c. The distance (d) is negative (-) if the point
P1 (x1 ,y1) is below the line.

Examples:
1. Find the distance from the line 5x = 2y + 6 to the
points
a. (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.

Lesson 6 straight line

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Lesson 6 straight line