# Thread: Locus of Points

1. ## 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?

2. 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)

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

thus answer is
$\displaystyle y=3x+1$

3. ## Very easy......

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)

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

thus answer is
$\displaystyle y=3x+1$
Thank you for the easy explanation and diagram.