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CHAPTER 3
THE STRAIGHT LINE
DR. SARMIENTO
(c) 1999
1. EQUATION OF THE LINE.-To determine the equation of a line we just need
two characteristic properties from which we:
i. Assume that the equation has the general form Ax + By C =0.
ii. Express the given conditions algebraically. This will generate two equations in
three unknowns A, B, and C.
iii. Solve these equations for two of the coefficients, A, B, or C, in terms of the third
one.
iv. Substitute back in the general form of the equation the values found in iii. and
divide by the third coefficient.
Example: Find the equation of the line that passes through the points (0,3) and (5,2).
Since the line contains the given points, the following equations are satisfied
A(0) + B(3) + C =0
A(5) + B(2) + C =0
Solving the first one for B we get B = - C/3 and substituting this value into the second
equation gives
Which is the desired equation.
In some instances it could be more convenient to use the slope y-intercept form y =
mx + b and, once the parameters m and b are found, we can express, if we wish, the
equation in general form.
From the previous example the parameters m and b can be found by solving the
system
3 = m(0) + b
2 = m(5) + b
from which b = 3 and m = -1/5. Substituting these values into the slope y-intercept
form of the equation we get
y = -1/5 x + 3
which is equivalent to
x + 5y -15 = 0.
2. POINT-SLOPE FORM.-The equation of the line that passes through the point
P(x1,y1) with slope m is
y - y1 = m(x - x1)
In effect, from the general form of the equation
Ax + By + C = 0 ----------- [1]
we have that
Ax1 + By1 + C = 0 --------- [2]
and also that
-A/B = m
Subtracting [2] from [1] we get
A(x - x1) + B(y - y1) = 0
and since A = -mB
-mB(x - x1) + B(y - y1) = 0
Finally, dividing by B and solving for y - y1 we get
y - y1 = m(x - x1)
Even in a simpler way the same result could be obtained if we use the slope y-intercept form y =
mx + b. From it
y1 = mx1 + b
and subtracting this from the previous one and factoring out m we get
y - y1 = m(x - x1)
This form allow us to find easily the equation of the line when a point and the slope are given,
reason for which is known as the point-slope form.
3. COROLLARY 1.-The equation of the line passing through the points P(x1,y1) and Q(x2,y2) is
In effect, by definition the slope of the line pasing through the points P and Q is
Substituting this value into the equation y - y1 = m(x - x1) gives the desired result.
Example: Find the equation of the line that passes through the points P(0,3) and Q(5,2).
Substituting the coordinates of the given points into the above equation gives
Result that is identical to the one obtained in the previous example.
4. COROLLARY 2.-The necessary and sufficient condition for three points, P1(x1,y1),
P2(x2,y2), and P3(x3,y3), to be collinear (in the same line) is that the third order
determinant
be equal to zero.
In effect, the given points will belong to the same line if the slope of the line defined
by P1 and P3 is equal to then slope of the line defined by P1 and P2, that is
Multiplying and regrouping terms we have
The left hand side of this equality corresponds to the expansion of the above
determinant.
5. COORDINATES AT THE ORIGIN.-If a and b are respectively, the abscissa and
the ordinate at the origin (x and y intercepts) of a straight line, its equation can be
expressed as
In effect, assuming the line has the form Ax + By + C = 0, since it passes through the
points (a,0) and (0,b) we have that
Aa + C = 0
and
Bb + C = 0
Solving for A and B and substituting these values back into the general equation we
get
Dividing by C and transposing terms we end up with
which is the desired result.
An alternate form of getting the same result is through the slope y-intercept form y =
mx + b. In effect, since the line passes through the points (a,0) and (0,b),,its slope
would be
Substituting this value into the equation y = mx + b we have
Dividing by b and transposing terms we get
as we had anticipated.
Exercise: Express the equation of the line 2x - 4y -12 = 0 in terms of its coordinates
at the origin (x and y intercepts), and determined their values.
Dividing both sides of the equation by 12 and transposing terms we get
Therefore the coordinates at the origin are a = 6 and b = -3; that is, the x and y
intercepts are (6,0) and (0,-3) respectively.
6. LINEAR COMBINATIONS.-The set of lines passing through the intersection of
two given ones A1x + B1y + C1 = 0 and A2x + B2y + C2 = 0 is given by the equation
A1x + B1y + C + k(A2x + B2y + C2) = 0, known as the linear combination of the given lines,
and in which k is an arbitrary constant.
In effect, if (x1,y1) is the intersection point of the two given lines then
A1x1 + B1y1 + C1 = A2x1 + B2y1 + C2 = 0
and therefore
A1x + B1y + C + k(A2x + B2y + C2) = 0
which proves that the linear combination passes through the intersection point of the given lines.
Using linear combinations we can determine specific lines of the system without finding the
coordiantes of the intersection point.
Example: Find the eqaution of the line that passes through the intersection point of 3x - 5y -10
= 0 and x + y + 1 = 0 and is parallel to 4x - 3y -7 = 0.
The equation we are looking for bellongs to thge linear combination
3x - 5y -10 + k(x + y + 1) = 0
that is
(3 + k)x + (k -5)y + (k - 10) = 0
and being parallel to 4x - 3y -7 = 0 it must be
Therefore the desired equation is
or, equivalently
32x - 24y - 59 = 0
EXERCISES
1. Find the equation of the line parallel to 4x - y - 2 = 0 which passes through the point (-1,3).
2. Determine if the points (2,3), (-4,6), and (8,12) are collinear.
3. Two adjacent sides of a parallelogram have equations 2x + 3y - 7 = 0 and x - 3y + 4 = 0. Find
the equations of the other two sides knowing that they intersect each other at (3.2).
4. Find the area, the intersection point, and the equation of the diagonals of the quadrilateral
defined by the equations y = 0, y = 1, x +2y - 2 = 0, and x + 2y - 6 = 0.
5. Using linear combinations find the equation of the line passing through the intersection point
of 2x + 3y = 0 and x - 2y + 5 = 0. Express the result in terms of the coordinates at the origin.