Respuesta :
Answer:
a. [tex]x=-2[/tex]
b. [tex]y=7[/tex]
c. [tex]y = 6x+c[/tex]
d. [tex]y=-\frac{1}{6}x+c[/tex]
Step-by-step explanation:
part a: parallel to x = 3
[tex]x = 3[/tex] it is a vertical line that passes through [tex]x = 3[/tex], so a parallel line to this that passes through (-2,7) must be [tex]x=-2[/tex] which is also a vertical line (thus they are parallel because they will never touch).
part b: perpendicular to x = -3
[tex]x = -3[/tex] it is a vertical line that passes through [tex]x = -3[/tex], so a perpendicular line to this that passes through (-2,7) must be[tex]y = 7[/tex] because this is horizontal line, and since one line is vertical and the other is horizontal they are perpendicular.
part c: parallel to y = 6x - 13
For two lines to be parallel they must have the same slope. The slope of a line is the number that accompanies the x. for [tex]y=6x-13[/tex] the slope is 6. the equation for a parallel line will be: [tex]y = 6x+c[/tex] (where c can be any number) because the slope (6) is the same for both.
part d: perpendicular to y=6x-13
The condition for two lines to be perpendicular is:
[tex]m_{1}m_{2}=-1[/tex]
where [tex]m_{1}[/tex] is the slope of one line, and [tex]m_{2}[/tex] the slope of the perpendicular line. We have the slope of the first line as we found in part c: [tex]m_{1}=6[/tex] so the slope of the second line has to be:
[tex]m_{1}m_{2}=-1[/tex]
[tex]6m_{2}=-1[/tex]
[tex]m_{2}=-\frac{1}{6}[/tex]
so an equation of a line with a slope of [tex]-\frac{1}{6}[/tex] can be:
[tex]y=-\frac{1}{6}x+c[/tex] (where c can be any number)
