### Functions: Lines and linear functions

### Equations and lines

Relative positions of one line to another

Two lines in the plane can be in three different positions relative to each other:

- they can cross (or: intersect in one point),
- they can be parallel but unequal, or
- they can be identical.

The equations of these lines with unknowns #x# and #y# have, in the three respective cases: one solution (**regular**), no solution (**contradicting**), and the same line as solution (**dependent**).

In all cases, the points which lie on both lines are the solution to the system of two equations.

The first case occurs then and only then if the *slopes* of these two lines are different.

The statements follow from the fact that the following three statements are equivalent:

- the point #\rv{a,b}# lies on the line #c\cdot x+d\cdot y=f#
- #x=a\land y=b# is a solution to the equation #cx+dy=f#
- #c\cdot a +d\cdot b=f#

In the dependent case the two original equations are equivalent if neither can be reduced to the trivial equation #0=0#.

The two lines #l# and #m# are given by the following linear equations:

\[ \begin{array}{lrcl}l:&\quad 4\cdot y+5\cdot x+4 &=& 0 \cr m:&\quad -y-x-4 &=& 0 \end{array}\]Determine the coordinates of the intersection point of the two lines. Give the answer in the form #x=a\land y=b# for the correct values of #a# and #b#.

\[ \begin{array}{lrcl}l:&\quad 4\cdot y+5\cdot x+4 &=& 0 \cr m:&\quad -y-x-4 &=& 0 \end{array}\]Determine the coordinates of the intersection point of the two lines. Give the answer in the form #x=a\land y=b# for the correct values of #a# and #b#.

#x= 12 \land y = -16#

We describe the substitution method to get to this solution.

\[ \begin{array}{rclcl} 4\cdot y+5\cdot x=-4 & \land &-y-x=4 &\phantom{xxx}&\color{blue}{\text{switch solutions if needed}}\\

&&&\phantom{xxx}&\color{blue}{\text{in such a way that }x\text{ occurs in the first equation}}\\

x=-{{4\cdot y}\over{5}}-{{4}\over{5}}& \land &-y-x=4 &\phantom{xxx}&\color{blue}{\text{express }x\text{ in }y\text{ in the first equation}}\\

x=-{{4\cdot y}\over{5}}-{{4}\over{5}} & \land & {{4}\over{5}}-{{y}\over{5}}=4 &\phantom{xxx}&\color{blue}{\text{replace }x\text{ by the expression in }y}\\

&& &\phantom{xxx}&\color{blue}{\text{from the first equation in the second one }}\\

x = -{{4\cdot y}\over{5}}-{{4}\over{5}} &\land & y = -16&\phantom{xxx}&\color{blue}{\text{solve the second equation}}\\

x = 12 &\land & y = -16&\phantom{xxx}&\color{blue}{\text{enter the found value for }y\text{ in}}\\

& & &\phantom{xxx}&\color{blue}{\text{in the first equation}}\\ \end{array}

\]

Hence, the coordinates of the intersection point of the lines #l# and #m# is #x= 12\land y = -16#.

We describe the substitution method to get to this solution.

\[ \begin{array}{rclcl} 4\cdot y+5\cdot x=-4 & \land &-y-x=4 &\phantom{xxx}&\color{blue}{\text{switch solutions if needed}}\\

&&&\phantom{xxx}&\color{blue}{\text{in such a way that }x\text{ occurs in the first equation}}\\

x=-{{4\cdot y}\over{5}}-{{4}\over{5}}& \land &-y-x=4 &\phantom{xxx}&\color{blue}{\text{express }x\text{ in }y\text{ in the first equation}}\\

x=-{{4\cdot y}\over{5}}-{{4}\over{5}} & \land & {{4}\over{5}}-{{y}\over{5}}=4 &\phantom{xxx}&\color{blue}{\text{replace }x\text{ by the expression in }y}\\

&& &\phantom{xxx}&\color{blue}{\text{from the first equation in the second one }}\\

x = -{{4\cdot y}\over{5}}-{{4}\over{5}} &\land & y = -16&\phantom{xxx}&\color{blue}{\text{solve the second equation}}\\

x = 12 &\land & y = -16&\phantom{xxx}&\color{blue}{\text{enter the found value for }y\text{ in}}\\

& & &\phantom{xxx}&\color{blue}{\text{in the first equation}}\\ \end{array}

\]

Hence, the coordinates of the intersection point of the lines #l# and #m# is #x= 12\land y = -16#.

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