[tex]\bf (\stackrel{x_1}{6}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{-6}~,~\stackrel{y_2}{4}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{4-4}{-6-6}\implies \cfrac{0}{-12}\implies 0 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-4=0(x-6)\implies y-4=0\implies y=4[/tex]
Answer: [tex]y-4=0[/tex]
Step-by-step explanation:
The equation of line passing through two points (a,b) and (c,d) is given by :-
[tex](y-b)=\dfrac{d-b}{c-a}(x-a)[/tex]
The standard form of equation of a line is given by :-
[tex]Ax+By+C=0[/tex], where A , B , and C are integers.
Then , the equation of line passing through two points K(6,4) and L(-6,4) is given by :-
[tex](y-4)=\dfrac{4-4}{-6-6}(x-6)\\\\\Rightarrow\ y-4=(0)(x-6)\\\\\Rightarrow\ y-4=0[/tex]
