Equation of right-hand side

Definition

The equation of a right ''' ''' $D$ is one (or several) equation has several unknown factors (coordinate), and whose whole of the solutions form the line $D$.

In the plan

In the plan, the whole of the points $M\; \backslash \; left\; (X,\; there\; \backslash \; right)$ formant $D$ can be represented by an equation of the form:

$ax\; +\; B\; there\; +\; C\; =\; 0$

where $a$, $b$ and $c$ are constants. In this case,

$D\; =\; \backslash \; \{(X,\; there)\; \backslash \; in\; \backslash \; mathbb\; \{R\}\; ^2\; \backslash \; |\; \backslash \; ax\; +\; by\; +\; C\; =\; 0\; \backslash \}$

In space

In a space with three dimensions, the whole of the points $M\; \backslash \; left\; (X,\; there,\; Z\; \backslash \; right)$ formant $D$ can be represented by a system of two equations of the form: $\backslash \; begin\; \{boxes\}\; ax+by+cz+d=0\; \backslash \; \backslash \; a\text{'}\; x+b\text{'}\; y+c\text{'}\; z+d\text{'}=0\; \backslash \; end\; \{boxes\}$

where $a$, $b$, $c$, $d$, $a\text{'}$, $b\text{'}$, $c\text{'}$, $d\text{'}$ are constants.

$ax+by+cz+d=0$ and $a\text{'}\; x+b\text{'}\; y+c\text{'}\; z+d\text{'}=0$ are two equations of plan.

Example

In the plan, the line $D$ passing by the points $A\; (-\; 1,4)$ and $B\; (1,0)$, has as an equation:

$-2x\; -\; there\; +\; 2\; =\; 0.$

Particular cases

In the plan, a line $D$ parallel with the x-axis (horizontal) has an equation of the form:

$there\; =\; y\_0.$ with $y\_0\; \backslash \; in\; \backslash \; mathbb\; \{R\}$

In the same way, a line $D$ parallel with the y-axis (vertical) has an equation of the form:

$X\; =\; x\_0.$ with $x\_0\; \backslash \; in\; \backslash \; mathbb\; \{R\}$

Seek of an equation of right-hand side

1) Characterization of an equation of right-hand side: That is to say the equation with two unknown factors there = 3x - 2. Seek 5 couples solutions of this equation. Represent in a reference mark the associated points.

Let us seek solutions.

It is necessary to choose a value for X then calculate the value of there corresponding. For example: If X = 0 then there = 3 X 0 - 2; there = -2 A couple solution is (0; -2)

If X = 1 then there = 3 X 1 - 2; there = 1 (1; 1) is solution of the equation

If X = 2 then there = 3 X 2 - 2; there = 4 I find the solution (2; 4).

If X = -1 then there = 3 X (- 1) - 2; there = -5 A couple solution is (- 1; -5)

if X = 1/2 then there = 3 X 1/2 - 2 = 3/2 - 4/2; there = -1/2. A couple solution is (1/2; -1/2)

Let us represent in a reference mark (O, I, J) these solutions on a graph by associating with each one of these couples a point which has the same coordinates.

All the points are aligned.

The equation with two unknown factors there = 3x - 2 is an equation of right-hand side.

Number 3 represents the slope of the right-hand side. It is the directing coefficient.

Number -2 represents the ordinate of the point of X-coordinate 0, intersection of the right-hand side with the y-axis. It is the ordinate in the beginning.

By resolution of a system of equations

Are two not confused points of the plan, $M\; \backslash \; left\; (U,\; v\; \backslash \; right)$ and $M\text{'}\; \backslash \; left\; (u\text{'},\; v\text{'}\; \backslash \; right)$.

If the line passing by these two points is not vertical ($u\; \backslash \; not=u\text{'}$), its equation is $y=ax+b$.

To find its equation, the system should be solved: $\backslash \; begin\; \{boxes\}\; v=au+b\; \backslash \; \backslash \; v\text{'}=au\text{'}+b\; \backslash \; end\; \{boxes\}$

There is $a=\; \backslash \; cfrac\; \{v\text{'}-v\}\; \{u\text{'}-u\}$ (directing coefficient).

To find the constant $b$ (ordered in the beginning), it is necessary to replace the variables $x$ and $y$ respectively by $u$ and $v$ (or $u\text{'}$ and $v\text{'}$).

There is then $v=a\; \backslash \; times\; u+b\; \backslash ,\; \backslash \; Leftrightarrow\; \backslash ,\; b=v-a\; \backslash \; times\; u$.

The equation of the right-hand side is then with final the $\backslash \; left\; (ME\; \backslash \; right):\; y=\backslash cfrac\{v\text{'}-v\}\{u\text{'}-u\}x+v-a\backslash times\; u$

By colinearity of two vectors

Are $A$ and $B$ two points not confused of the plan.

$M\; \backslash \; left\; (X,\; there\; \backslash \; right)$ is a point of the right-hand side $\backslash \; left\; (AB\; \backslash \; right)$ if and only if the vectors $\backslash \; overrightarrow\; \{AB\}$ and $\backslash \; overrightarrow\; \{AM\}$ are colinéaires.

There one obtains the equation of the right-hand side by writing $x\_\; \{\backslash \; overrightarrow\; \{AB\}\}\; y\_\; \{\backslash \; overrightarrow\; \{AM\}\}\; \_\; \{\backslash \; overrightarrow\; \{AB\}\}\; x\_\; \{\backslash \; overrightarrow\; \{AM\}\}\; =0$.

I.e. $\backslash \; left\; (x\_B-x\_A\; \backslash \; right)\; \backslash \; left\; (there\; there\; \_A\; \backslash \; right)\; -\; \backslash \; left\; (y\_B\; there\; \_A\; \backslash \; right)\; \backslash \; left\; (x-x\_A\; \backslash \; right)\; =0$.

Remarks

• a line can have an infinity of equations which represents it.

• In the plan, any line admits an equation (known as Cartesian) form: $ax\; +\; by\; +\; C\; =\; 0$.

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