# Thread: Find the point on line...That is equidistant from...

1. ## Find the point on line...That is equidistant from...

Hi My question is:

Find the point on the line 4x-2y+3=0 that is equidistant from (3,3) and (7,-3)

I'm still a little confused about what to do.

Some ideas: I set 4x-2y+3=0 to y=2x + 3/2
and that gives me the y-intercept

Now, I was thinking that I need to find x with a distance formula.

For example, find the distance formula for (3,3), then a second distance formula for (7,-3) and equal them to each other. Plugging in the y-intercept, then solving for x. I tried that, but I had an odd answer.

If anybody could help me out that would be great! thanks

2. Originally Posted by kn336a
Hi My question is:

Find the point on the line 4x-2y+3=0 that is equidistant from (3,3) and (7,-3)

I'm still a little confused about what to do.

Some ideas: I set 4x-2y+3=0 to y=2x + 3/2
and that gives me the y-intercept

Now, I was thinking that I need to find x with a distance formula.

For example, find the distance formula for (3,3), then a second distance formula for (7,-3) and equal them to each other. Plugging in the y-intercept, then solving for x. I tried that, but I had an odd answer.

If anybody could help me out that would be great! thanks
See your question is just a use of disance of a point from line

the distance of a point
$(x_1,y_1)$ from line $ax+by+c=0$ is given by

$
|{\frac{ax_1+by_1+c} {\sqrt{a^2+b^2}}}|
$

Use it to find and equate those distances

3. Originally Posted by kn336a
Find the point on the line 4x-2y+3=0 that is equidistant from (3,3) and (7,-3)
Find the equation of the perpendicular bisector of the line segment with endpoints (3,3) and (7,-3).
Its slope must be $\frac{2}{3}$ and contain the midpoint $(5,0)$.
Using that line find its intersection with the given line.
That point is the answer to this question.

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# find the point on the line that is equidistant

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