# Locus of Points

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• Feb 10th 2009, 09:09 PM
magentarita
Locus of Points
Find an equation that represents the locus of points equidistant from the lines whose equations are y = 3x + 8 and y = 3x - 6?
• Feb 10th 2009, 09:44 PM
ADARSH
For equations like
y= mx+ c

c is the y- intercept (value of y at x=0)

y = 3x + 8 and y = 3x - 6

thus y interceept of our two lines are 8 and (-6)

So if we want a point on y axis equidistant from these lines
we need to have a value of y-intercept in between them

-- this means the value of y at x=0 of our answer should be in between 8 and (-6)

And the line should be parralel so that its equidistant throughout (see the attatchment)

$
y=3x+\frac{8+(-6)}{2}

$

thus answer is
$y=3x+1$
• Feb 12th 2009, 06:06 PM
magentarita
Very easy......
Quote:

Originally Posted by ADARSH
For equations like
y= mx+ c

c is the y- intercept (value of y at x=0)

y = 3x + 8 and y = 3x - 6

thus y interceept of our two lines are 8 and (-6)

So if we want a point on y axis equidistant from these lines
we need to have a value of y-intercept in between them

-- this means the value of y at x=0 of our answer should be in between 8 and (-6)

And the line should be parralel so that its equidistant throughout (see the attatchment)

$
y=3x+\frac{8+(-6)}{2}

$

thus answer is
$y=3x+1$

Thank you for the easy explanation and diagram.