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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