Analytical Geometry simplified method for parallel and perpendicular lines

As a professor of mathematics at all levels of education by the continued practice and use purely logical reasoning and a simple analysis of the various topics of this science I could deduct a myriad of methods and practical ways to solve different mathematical situations in a simple manner without resorting to cumbersome even sometimes formulas, theorems or properties that confuses their understanding of all students in these subjects.

For this article I have chosen at random within the Analytical Geometry analysis of parallel and perpendicular lines on time, but not the whole subject in itself, because my intention is to show only a topic of this for obvious reasons. In future articles I will expose other points of interest using this SIMPLE WAY TO SEE THE MATH? STICS.

Before continuing I must emphasize that this article is intended for connoisseurs of this matter, who have experience in this, in particular should not be considered by students who are new in this matter not to give rise to confusion over who have at least This knowledge base can analyze it carefully.

THE EQUATION OF THE STRAIGHT: From the cumbersome theory is known that the equation of the line consists of points of the set of ordered pairs (X, Y), in the Cartesian plane defined by the axes X and Y obey a rule training given in various ways, including for example:

Y = mX + b or aX + bY + c = 0

Where for a value of X will correspond a value to Y. Both lines are plotted in the X-Y. The value of "m" is the gradient of the line on the X axis and "m" is related to the second equation by m =-a / b. It is said that 2 lines are parallel if both have the same slope and are perpendicular when the product? Of their slopes gives us the factor -1.

To analyze the case of parallel and perpendicular lines in a simple and go directly to analyze these cases by one problem in each case.

1) .- PARALLEL LINES:

PROBLEM: You have the line L1: 3x + 4y +7 = 0, we want to find the equation of the line L2 parallel to L1 so L2 passes through the point A (3, -2).

SOLUTION: As 2 parallel lines have the same slope m =-a / b (aX + bY + c = 0) in our case L1: have slope m = -3 / 4 as well as L2: m = -3 / 4 with we can conclude that if L1 is: 3x + 4y + 7 = 0, L2 also have their first 2 to 3X 4Y factors but will be another independent term, ie L2: 3X + 4Y + K = 0. Osea:

L2: 3X + 4Y + K = 0 is parallel to 3x + 4y + 7 = 0

And as in the problem data tells us that the point A (3. -2) Belongs to the line L2, then, to find the equation we would fail to find the requested value is obtained by replacing K in L2:

L2: 3 (3) + 4 (-2) + K = 0 K = -1 so:

L2: 3x + 4y - 1 = 0

Normally the solution of this problem would have led to a series of formulas and theorems longer. What we have done in this case was only that bearing in mind that if a line has the form aX + bY + k1 = 0, the line parallel to it shall take the form aX + bY + k2.



PERPENDICULAR LINES:

PROBLEM: You have the line L1: 2x + 3y + 11 = 0 is desired to find the equation of the line L2 perpendicular to L1 so L2 passes through the point A (1.2).

SOLUTION: For 2 perpendicular lines marks the product of their slopes is -1, ie m1.m2 = m1 = -1 or -1/m2 ie if the slope of the first line is m1 = a / b, the slope of the second will be m2 = - b / a is the negative of the reverse split, meaning that if L1 is: 2x + 3y + 11 = 0 then the equation of L2 is: 3X - 2Y + K = 0. In order not to confuse us so we can generalize it by reasoning that if L1: aX + bY + c = 0 the equation of perpendicular line is: L2: bX - aY + K = 0. Then:

L2: 3X - 2Y + K = 0 is perpendicular to:

L1: 2x + 3y + 11 = 0

And the problem of data L2 we are told through the point A (1.2) To find the equation of the line would fail us find the value of replacing K in L2

L2: 3 (1) - 2 (2) + K = 0 K = 1 so:

L2: 3x - 2y + 1 = 0.

Just as in the previous market, we have solved this problem more easily than traditional methods and mechanics. For this second case, we conclude that: L1: aX + bY + K1 = 0 perpendicular to the line has the form L2: bX - aY + K2 = 0.