Geometry of the plan
1) Equations that describe the straight ones in the plan:
The straight ones in the plan can be described in 2 various ways:
Explicit cartesian equation
Draft of the shape ax1 bx2 = c where:
x1 = abscissa of a whichever point of the straight one
x2 = ordered associated to x1 for that straight one
c = ordered to the origin
From this it is passed to the explicit shape
easily dove 
where - b/a = m come defined angular coefficient in how much if the famous origin is supposed straight passing per that x2/x1 are characteristic a constant relationship for that straight one, while if whichever is taken one other straight one, such varied relationship but will continue to being constant for that straight one.
2) Straight parallels:
Based on how much as soon as said it succeeds intuitivo to think that 2 straight ones are parallels when they only differ for the coordinate to origin c , while the relationship - b/a it remains immutato.
3) Straight orthogonal:
Straight r2 are orthogonal to straight rthe 1 when with it form an angle of 90°.
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Observing the design if 1 is placed then ob = m1 = oa/ob = oa and also m2 = oc/ob = oc. But Euclide says to us that if the triangle is rectangle then ab * bc = 12 | m1 * m2 | = 1 from which it follows that the angular coefficient of a straight one is equal to the inverse one of the angular coefficient of the straight one to orthogonal it and opposite sign inasmuch as sure falls in an other quadrant of the plan. As an example:
r1 5x1 2x2 = 2 m = -5/2
is orthogonal to the straight one
r2 5x1 - 2x2 = c m = 2/5
More in a generalized manner, it has been
confusion source and it will continue it are it, we can describe one
orthogonal carrierv ^ to
straight rthe 1 and passing for
the such origin that:
v^ = 
and one carrier v// parallel to straight: the
v// = 
All that drift from the observation that if we take 2 punti on straight rthe 1 , x and y then their difference is a carrier parallel to straight rthe 1 applied in the origin. If of this carrier (x - y) we make the product to climb with v^ we obtain 0 and therefore the 2 carriers are between of orthogonal they. If instead we make the product to climb with carrier v// we obtain the module of (x-y) therefore the 2 carriers are parallels.
** Ricordiamo in fact that the product to climb of 2 carriers is not that the orthogonal projection of one of the 2 on straight dove the lies the other; such projection is not a carrier but a number and is characterized from the product of the norms of the 2 carriers for the angle between they subtended.
3) parametric Equation:
Draft of the shape
= 
t
where:
= coordinated of the generic point of
the straight one
= coordinated of a point
whichever on the straight one
= coordinated of a tried
carrier parallel to the straight one and applied in the origin of the
plan
t = parameter alcui to vary
we are in a position to making to assume to the
value of all the points on the straight one.
In practical this form to it is based on the observation that a straight one can be described having to disposition a point for which it passes and a carrier to it parallel passing for the origin of the plan.
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From this famous figure easily that generic point x is found executing the vectorial sum between carrier v multiplied for one parameter simply t and the carrier op , vien from if that to varying of (t * v) we are in a position to describing whichever point x on retta rthe 1 .
3) Straight parallels with the parametric equation:
Based on how much as soon as said it succeeds intuitivo to think that 2 straight ones are parallels when have the same carrier director v or one to it proporziona them. As an example:
r1 = 
t
is parallel to retta the
r2 = 
t
4) Straight orthogonal with the parametric equation:
Straight r2 are orthogonal to straight rthe 1 when with it form an angle of 90°.
Also it is worth the same observation made for the cartesian equation here, in particular we must characterize one carrier v1 that is orthogonal to carrier v2
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It returns to be worth how much asserted previously that is a straight one r2 for being otogonale to r1 must have the coordinates of the carrier the equal to inverse of the coordinates of carrier the director v1 and ordered director v2 of opposite sign inasmuch as sure it falls in an other quadrant of the plan.
As an example:
r1 = t is orthogonal to retta the r2 = t
5) solvable Problems easily with the parametric equation:
a) Straight passing for 2 distinguished points p1 and p2
In order to identify a straight one with the parametric equation is necessary to us:
1 punto we can take one of 2 p 1as an example
1 carrier passing for the origin and parallel to the straight one we can take to the carrier difference that is p1 - p2
therefore we have
characterized the parametric equation of the straight one that to
varying of t describes all the points of retta the same
. As an example they are p1 = and p1 = 
then carrier v is worth - =
= v
therefore the complete
parametric equation turns out to be:
= t
b) Given one straight ra 1 and external point p to it to find the equation of the straight one passing for p orthogonal to r1
It is resolved to the usual considering that they are necessary to us:
punto we can take p
1 carrier passing for the origin and parallel to straight the carrier we take it directly orthogonal to the straight given using demonstrated how much already.
therefore if they are dati: p =
e
=
t
then tried the straight equation of the
orthogonal one is worth: = t
.
c) Intersection between straight 2 r1 and r2 of given parametric equation
It is obtained easily imposing that the generic
point belongs to both the straight ones, this gives
place to a system composed from 2 equations in the incognito t and s
from which these can be deduced then 2 parameters, replacing or s or t
in the respective equation we must find the same searched point
representative the intersection of the 2 straight ones. The
following eventualities are possible:
c1) No solution if the two straight ones are parallels
c2) a brace of values (x1 , x2) if the two straight ones are intersected in a single point
c3) Infinite solutions if the straight ones are coinciding, this happens as an example when one of the 2 parameters is free
As an example:
= t
e
= s
the generic points are:
e
from which:
=
whose
solutions are: t = 0 and
s = 0
therefore the encounter point is p = .
c4) Given one straight ra 1 and point p to calculate the distance of p respect to r1
It is obtained which elaboration of previous in 2 steps succeeded to you:
a) is estimated the straight equation of the orthogonal one to r1 passing for p
b) finds the point q of intersection between the 2 straight ones
c) is estimated the distance of p from q through the theorem of Pitagora that is:
6) Passage from one cartesian equation to the correspondent parametric equation:
2 points are found on the straight one and then algorithm 1 for the search of the parametric equation of the straight passing for 2 points is applied.
7) Passage from one parametric equation to the correspondent cartesian equation:
the 2 coordinated of the generic point expressed from the parametric equation are taken and island t in both dopodichè uguagliando them is removed t and remained one equation in x1 and x2 , that one are the cartesian equation of the straight one.
8) the cartesian equation for the typical problem solving:
a) Straight passing for 2 distinguished points p1 and p2
is estimated to leave from the
coordinates of punti: the p1 =
and p2 = 
realizing the difference of the coordinates reported to cartesian equation ax1 bx2 = c
b) Intersection between 2 straight ones
It is obtained easily putting to system the equations in explicit shape, and resolving the system as an example with the method of Gauss reducing it to a matrix, the solution will be following:
b1) No solution if the two straight ones are parallels
b2) a brace of values (x1 , x2) if the two straight ones are intersected in a single point
b3) Infinite solutions if the straight ones are coinciding, this happens as an example when one of the 2 incognito is free
c) Given one straight ra 1 and point p to calculate the distance of p respect to r1
It is obtained which elaboration of previous in 2 steps succeeded to you:
c1) having the cartesian equation of straight r1 ax1 bx2 = c passes to the correspondent parametric equation of the straight perpendicular remembering that:
one
orthogonal carrierv ^
to straight rthe 1 and
passing for the such origin that:
v^ =
therefore the equation è:
= t
c2) finds the point q of intersection between the 2 straight and the value of the correspondent parameter t
c3) is estimated the distance of p from q through the theorem of Pitagora that is:
