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## #1 2006-10-25 07:02:16

unique
Member
Registered: 2006-10-04
Posts: 419

### slope

find the equation of the line that is equidistant from the points (3,2) and(-4,-3)
write the equation in slope intercept form.

ok what is equidistant?

Desi
Raat Key Rani !

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## #2 2006-10-25 08:54:54

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

### Re: slope

You want to find the equation of the line that's made up of all the points that are equidistant from (3,2) and (-4,-3).

This will be the perpendicular bisector of the line between those 2 points.

The gradient between those 2 points is (2-(-3))/(3-(-4)) = 5/7.
The mid-point of the line is (-1/2,-1/2), by averaging each of the two co-ordinates.

So the perpendicular bisector will go through (-1/2,-1/2) and have a gradient of -7/5, so that it is perpendicular.

We can work out the y-intercept as being -1/2 - (-1/2*-7/5) = -6/5

Therefore, the equation of your line is y = -7/5x - 6/5. Or -5y = 7x+6 if you prefer.

Why did the vector cross the road?
It wanted to be normal.

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## #3 2006-10-26 03:41:05

unique
Member
Registered: 2006-10-04
Posts: 419

### Re: slope

THANKS for the help

Desi
Raat Key Rani !

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