Answer:
[tex]\mathrm{1.}\; m=\mathrm{N/A};\: \mathrm{none}; \: x=14\\\mathrm{2.}\; m=0;\: (0,-5); \:y=-5\\\mathrm{3.}\; m=2;\:(0, -6); \: -2x+y=-6\\[/tex]
Step-by-step explanation:
Problem 1:
- The slope of a vertical line is always undefined.
- The y-intercept of a line occurs when [tex]x=0[/tex]. In this line, [tex]x[/tex] is defined already as [tex]14[/tex] and therefore there is no y-intercept.
- The equation of a vertical line is [tex]x=n[/tex], where [tex]n[/tex] is some constant.
Problem 2:
- The slope of a horizontal line is always [tex]0[/tex].
- The y-intercept of a line occurs when [tex]x=0[/tex]. In this line, [tex]y[/tex] is declared as 5 for [tex]x\in \mathrm{R}[/tex]. Therefore, the y-intercept is [tex](0, -5)[/tex].
- The equation of a horizontal line is [tex]y=n[/tex], where [tex]n[/tex] is some constant.
Problem 3:
- The slope of a line given two points can be found using [tex]m=\frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1}[/tex]. In this case, the slope is [tex]\frac{11-(-1)}{5-(-1)}=\frac{12}{6}=2[/tex].
- The y-intercept occurs when [tex]x=0[/tex]. Using the equation we found (refer below), plug in [tex]x=0[/tex] to get the y-coordinate of the y-intercept.
- In slope-intercept form, the equation of a line is written as [tex]y=mx+b[/tex]. Using [tex]m=2[/tex] and any coordinate this line passes through, we find the equation of this line to be [tex]y=2x-6[/tex]. Rearranging this in standard form ([tex]ax^2+bx+c[/tex]), we get [tex]-2x+y=-6[/tex].
