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PARALLEL AND PERPENDICULAR LINES The
general equation of a straight line is often written with capital letters for
coefficients, as follows: Ax+By+C=0 These
literal coefficients, as they are called, represent the numerical coefficients
encountered in a typical linear equation. Suppose
we are given two equations that are duplicates except for the constant term, as
follows: Ax +By+C=0 Ax +By+D=0 By
placing these two equations in slopeintercept form, we can show that their slopes
are equal, as follows: y =
Thus,
the slope of each line is A/B. Since
lines having equal slopes are parallel, we reach the following conclusion: In any two linear equations, if the coefficients
of the x and y terms are identical in value and sign, then the lines
represented by these equations are parallel. EXAMPLE: Write the equation of a line parallel to 3x  y  2 = 0 and passing
through the point (5,2). SOLUTION: The coefficients of x and y in the desired equation
are the same as those in the given equation. Therefore, the equation is 3xy+D=0 Since
the line passes through (5,2), the values x = 5 and y = 2 must satisfy the
equation. Substituting these, we have 3(5)
 (2) + D = 0 D= 13 Thus,
the required equation is 3xy
13 =0 A
situation similar to that prevailing with parallel lines involves perpendicular
lines. For example, consider the equations Ax
+By+C=0 BxAy+D=0 Transposing
these equations into the slopeintercept form, we have y =
Since
the slopes of these two lines are negative reciprocals, the lines are
perpendicular. The
conclusion derived from the foregoing discussion is as follows: If a line is to be perpendicular to a given line, the
coefficients of x and y in the required equation are found by interchanging the
coefficients of x and y in the given equation and changing the sign of one of
them. EXAMPLE: Write the equation of a line perpendicular to the line x + 3y + 3 = 0
and having a y intercept of 5. SOLUTION: The required equation is 3xy+D=0 Notice
the interchange of coefficients and the change of sign. At the point where the
line crosses the Y axis, the value of x is 0 and the value of y is 5.
Therefore, the equation is 3(0)  (5) + D = 0 D=5 The
required equation is 3xy+5=0 119 PRACTICE
PROBLEMS: 1.
Find the equation of the line whose perpendicular forms an angle of 135' from
the positive side of the X axis and whose perpendicular distance is V^{}2^{
}units from the origin. Find
the equations of the following lines: 2.
Through (1,1) and parallel to 5x  3y = 9. 3.
Through ( 3,2) and perpendicular to x + y = 5. ANSWERS:

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