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by Brent@GMATPrepNow » Sun Aug 11, 2013 5:24 pm
[email protected] wrote:Find a point on the Y axis which is equidistant from the points A(6,5) and B(-4, 3).
We'll use the following formula:
The distance between the points (a,b) and (c,d) = sqrt[(a-c)² + (b-d)²]

IMPORTANT: if we're looking for a point on the y-axis, then the x-coordinate of that point will equal 0. So, let's say that (0,y) is the point that's equidistant from points A and B.

We want: the distance from point A to (0,y) = the distance from point B to (0,y)
In other words, the distance from (6,5) to (0,y) = the distance from (-4, 3) to (0,y)
Apply formula: sqrt[(6 - 0)² + (5 - y)²] = sqrt[(-4 - 0)² + (3 - y)²]
Square both sides: (6 - 0)² + (5 - y)² = (-4 - 0)² + (3 - y)²
Expand and simplify: 36 + 25 - 10y + y² = 16 + 9 - 6y + y²
Simplify: 61 - 10y + y² = 25 - 6y + y²
Simplify: 36 = 4y
Solve: y = 9

So, [spoiler](0, 9)[/spoiler] is the point on the y-axis that's equidistant from the points A(6,5) and B(-4, 3).

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by [email protected] » Sun Aug 11, 2013 5:35 pm
But Brent, the point is equidistant doesn't it imply that the point is in the middle of both Y and X,

So cant we use the midpoint formula for this?

Brent@GMATPrepNow wrote:
[email protected] wrote:Find a point on the Y axis which is equidistant from the points A(6,5) and B(-4, 3).
We'll use the following formula:
The distance between the points (a,b) and (c,d) = sqrt[(a-c)² + (b-d)²]

IMPORTANT: if we're looking for a point on the y-axis, then the x-coordinate of that point will equal 0. So, let's say that (0,y) is the point that's equidistant from points A and B.

We want: the distance from point A to (0,y) = the distance from point B to (0,y)
In other words, the distance from (6,5) to (0,y) = the distance from (-4, 3) to (0,y)
Apply formula: sqrt[(6 - 0)² + (5 - y)²] = sqrt[(-4 - 0)² + (3 - y)²]
Square both sides: (6 - 0)² + (5 - y)² = (-4 - 0)² + (3 - y)²
Expand and simplify: 36 + 25 - 10y + y² = 16 + 9 - 6y + y²
Simplify: 61 - 10y + y² = 25 - 6y + y²
Simplify: 36 = 4y
Solve: y = 9

So, [spoiler](0, 9)[/spoiler] is the point on the y-axis that's equidistant from the points A(6,5) and B(-4, 3).

Cheers,
Brent
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by Brent@GMATPrepNow » Sun Aug 11, 2013 5:45 pm
[email protected] wrote:But Brent, the point is equidistant doesn't it imply that the point is in the middle of both Y and X,

So cant we use the midpoint formula for this?
The midpoint between A and B will, indeed, be equidistant from A and B. However, the midpoint will not be on the y-axis, and the question tells us that the point is on the y-axis.

The truth is that there is an infinite number of points that are equidistant from the points A(6,5) and B(-4, 3). To get a better idea of what I mean, check out this related post: https://www.beatthegmat.com/dint-get-the ... 67535.html

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Brent
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by mgm » Mon Aug 12, 2013 5:07 pm
Brent I did get the same answer but then I thought abt an alternative option as well: the point is below the x axis.

So (y+5)^2 + 36 = (y+3)^2 + 16

You get a different solution

Where is this problem from?
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by Brent@GMATPrepNow » Mon Aug 12, 2013 7:09 pm
mgm wrote:Brent I did get the same answer but then I thought abt an alternative option as well: the point is below the x axis.

So (y+5)^2 + 36 = (y+3)^2 + 16

You get a different solution

Where is this problem from?
How did you get (y+5) and (y+3)?
We are finding the distance from (6,5) to (0,y) = the distance from (-4, 3) to (0,y)

So, the equation will be EITHER sqrt[(6 - 0)² + (5 - y)²] = sqrt[(-4 - 0)² + (3 - y)²]
OR sqrt[(0 - 6)² + (y - 5)²] = sqrt[(0 - -4)² + (y - 3)²]

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by mgm » Tue Aug 13, 2013 3:42 am
Brent,

If you want to find a point below the X Axis that is equidistant from the two points and get the distance from say (0,-y) and the two points then you get a different answer...
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by Brent@GMATPrepNow » Tue Aug 13, 2013 6:35 am
mgm wrote:Brent,

If you want to find a point below the X Axis that is equidistant from the two points and get the distance from say (0,-y) and the two points then you get a different answer...
When we say that the point (0, y) is the point equidistant from points A and B, we are already allowing for the possibility that y may be negative. But it turns out that y is positive (y = 9)

Every point on the red line is equidistant from points A and B.
Image
The question asks us to find a particular point on the red line. It asks us to find the point that's on the y-axis (where x = 0).
The coordinates of that point as (0, 9)
As you can see from the graph, there are no other points on the y-axis that are equidistant from A and B.

In your approach, you let (0, -y) be the equidistant point. All you're doing there is saying "let -y be the point on the y-axis that is equidistant from A and B." That's fine, BUT when you solve the equation, you must solve it for -y. When you do, you get -y = 9, which means (0, 9) is the equidistant point (great). HOWEVER, if you solve it for y, you get y = -9, which is not correct.

I hope that helps.

Cheers,
Brent
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