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Probably working too hard, cause this should be easy... - HSN forum

# Probably working too hard, cause this should be easy...

### #1Helena

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Posted 18 May 2009 - 07:46 PM

Question in a REVISION PACK that I just can't do and it's driving me mad.
Find the equation of the straight line which passes through the point (-2,6) and is parallel to the line with equation x=3.

If x=3, then the gradient of the line I'm supposed to find the equation of is undefined, right? Or am I hallucinating or something...
Anyways, any help would be much appreciated.
Thanks.
xXx

### #2SncZ

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Posted 18 May 2009 - 08:37 PM

QUOTE (Helena @ May 18 2009, 07:46 PM) <{POST_SNAPBACK}>
Question in a REVISION PACK that I just can't do and it's driving me mad.
Find the equation of the straight line which passes through the point (-2,6) and is parallel to the line with equation x=3.

If x=3, then the gradient of the line I'm supposed to find the equation of is undefined, right? Or am I hallucinating or something...
Anyways, any help would be much appreciated.
Thanks.
xXx

yea gradients of vertical lines are undefined.

The Standard equation of a vertical line is ' x=a ' where a is the x-intercept. Since its a vertical line, x stays constant throughout.

Therefore the equation of your line is x = -2.

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