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EQUATION OF A STRAIGHT LINE In Mathematics, Volume 1, equations such as 2x+y=6 are designated as linear equations, and their graphs are
shown to be straight lines. The purpose of this discussion is to study the
relationship of slope to the equation of a straight line. POINTSLOPE FORM Suppose that we want to find the equation of a straight
line that passes through a known point and has a known slope. Let (x,y)
represent the coordinates of any point on the line, and let (x,,y,) represent
the coordinates of the known point. The slope is represented by m. Recalling the formula defining slope in terms of the
coordinates of two points, we have
EXAMPLE: Find the equation of a line passing through the
point (2,3) and having a slope of 3. SOLUTION:
The pointslope form may be used to find the
equation of a line through two known points. The values of x,, x_{2},
y,, and Y2 are first used to find the slope of the line; then
either known point is used with the slope in the pointslope form. EXAMPLE: Find the equation of the line through the points (
 3,4) and (4,  2). SOLUTION:
Letting (x,y) represent any point on the line and using (
 3,4) as
Using (4,  2) as the known point will also give 7y + 6x =
10 as the linear equation. SLOPEINTERCEPT FORM Any line that is not parallel to the Y axis intersects the
Y axis at some point. The x coordinate of the point of intersection is 0,
because the Y axis is vertical and passes through the origin. Let the y
coordinate of the point of intersection be represented by b.
Then the point of intersection is (O,b), as shown in figure 18. The y
coordinate, b, is called the y intercept. The slope of the line in figure 18 is
The value of Ay in this expression is y  b, where y
represents the y coordinate of any point on the line. The value of
x is x  0
= x, so
Figure 18.Slopeintercept form. This is the standard slopeintercept form of
a straight line. EXAMPLE: Find the equation of a line that intersects the Y
axis at the point (0,3) and has a slope of 513. SOLUTION:
PRACTICE PROBLEMS: Write equations for lines having points and slopes as
follows:
NORMAL
FORM Methods
for determining the equation of a line usually depend upon some knowledge of a
point or points on the line. Let's now consider a method that does not require
advance knowledge concerning any of the line's points. All that is known about
the line is its perpendicular distance from the origin and the angle between
the perpendicular and the X axis, where the angle is measured counterclockwise
from the positive side of the X axis. In
figure 19, line AB is a distance p away from the origin, and line OM forms an
angle 9 (the Greek letter theta) with the X axis. We select any point P(x,y) on
line AB and develop the
Figure 19.Normal form. equation
of line AB in terms of the x and y of
P. Since P represents ANY point on the line, the x and y of the equation will
represent EVERY point on the line and therefore will represent the line itself. PR is constructed perpendicular to OB at
point R. NR is drawn parallel to AB, and PN is parallel to OB. PS is perpendicular to NR and to AB.
A right angle is formed by angles NRO and
PRN. Triangles ONR and OMB are similar
right triangles. Therefore, angles NRO and
MBO are equal and are designated as
8'. Since 8 + 8' = 90' in triangle OMB and
angle NRO is equal to 8', then angle PRN equals 8. Finally, the x distance of
point P is equal to OR, and the y
distance of P is equal to PR. To
relate the distance p to x and y, we reason as follows:
This
final equation is the normal form. The word "normal" in this usage
refers to the perpendicular relationship between OM and AB. "Normal"
frequently means "perpendicular" in mathematical and scientific
usage. The distance p is always considered to be positive, and 8 is any angle
between 0' and 360'. EXAMPLE: Find the equation of the line that is 5 units away from the origin, if
the perpendicular from the line to the origin forms an angle of 30' from the
positive side of the X axis. SOLUTION.
