Finding the equation of a perpendicular line

errequeerre

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Jun 18, 2013
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16
Hello,

I had this problem to solve:
Find the equation of the perpendicular to the line y = 3 and passing through the point (4, 5).

I was baffled when I saw the question, because I can't make head or tail of it.
  • As I understand it, the line for an y = a equation is a horizontal line (slope zero).
  • Therefore, a perpendicular line must be a vertical line (undefined slope).
  • If a line is vertical, the value is just x = b, but it does not cross y, unless it is x = 0, in which case it coincides with y.
If (4, 5) means that it crosses x at 4 and y at 5, how can it be a perpendicular to the horizontal y = 3? :rolleyes:

Thanks for your help!
 
I had this problem to solve:
Find the equation of the perpendicular to the line y = 3 and passing through the point (4, 5).

Hint: \(\displaystyle x=4\) is a vertical line. Is \(\displaystyle (4,5)\) on that line?
 
Hello,

I had this problem to solve:
Find the equation of the perpendicular to the line y = 3 and passing through the point (4, 5).

I was baffled when I saw the question, because I can't make head or tail of it.
  • As I understand it, the line for an y = a equation is a horizontal line (slope zero). Yes
  • Therefore, a perpendicular line must be a vertical line (undefined slope). Yes
  • If a line is vertical, the value is just x = b, but it does not cross y, unless it is x = 0, in which case it coincides with y. If x = 0, the resulting line coincides with the y axis. But there are an infinite number of lines parallel to the y-axis.
If (4, 5) means that it crosses x at 4 and y at 5, how can it be a perpendicular to the horizontal y = 3? :rolleyes:

Thanks for your help!
.
 
Hello,

I had this problem to solve:
Find the equation of the perpendicular to the line y = 3 and passing through the point (4, 5).

I was baffled when I saw the question, because I can't make head or tail of it.
  • As I understand it, the line for an y = a equation is a horizontal line (slope zero).
  • Therefore, a perpendicular line must be a vertical line (undefined slope).
  • If a line is vertical, the value is just x = b, but it does not cross y, unless it is x = 0, in which case it coincides with y.
  • I think a large part of your problem is indicated where you say "it does not cross y". What do you mean by "y"? I suspect you mean the y- axis, which is the line "x= 0", but that has nothing to do with this problem.
If (4, 5) means that it crosses x at 4 and y at 5, how can it be a perpendicular to the horizontal y = 3? :rolleyes:
Again, "crosses x at 4" and "crosses y at 5" make no sense. what do you mean by "x" and "y" here? In any case, "(4, 5) doesn't mean that the line crosses anything. it means that it contains the point (4, 5) which is the point 4 places to the right of the y-axis and 5 steps above the x-axis. The line y= 3 is horizontal and a line perpendicular to that must be horizontal. It must be of the form x= a for some a. If the line x= a contains the point (4, 5), what must a be?

Thanks for your help!
 
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