# Thread: points on axis

1. ## points on axis

i've been set the question:

find the coordinates of the point on the y-axis which is equidistant from points A and D. point A is (0, 4) and D is (10, 6). i'm not sure how to go about figuring that one out. if any one could show me a step by step way of doing that, it would be appreciated.

thanks, Mark

2. Originally Posted by mark
i've been set the question:

find the coordinates of the point on the y-axis which is equidistant from points A and D. point A is (0, 4) and D is (10, 6). i'm not sure how to go about figuring that one out. if any one could show me a step by step way of doing that, it would be appreciated.

thanks, Mark
Mods please delete this post - I realised how dumb my answer was after I posted it...

3. do you mean add them together and divide by 2? because i'd only get 5. and thats not the answer given in the book. the answer is (0, 30) so i don't know how to get that

4. Originally Posted by mark
i've been set the question:

find the coordinates of the point on the y-axis which is equidistant from points A and D. point A is (0, 4) and D is (10, 6). i'm not sure how to go about figuring that one out. if any one could show me a step by step way of doing that, it would be appreciated.

thanks, Mark
The locus of all points equidistant to two fixed points is the perpendicular bisector of the distance between the two fixed ponts.

I've attached a rough sketch of the situation. You have to perform the following steps:

1. Coordinates of the midpoint of AD.
2. Slope of AD
3. Perpendicular slope of AD
4. Equation of the line passing through the midpoint perpendicular to AD
5. Point of intersection of this line and the y-axis

5. thanks earboth, that was a good explanation

6. Originally Posted by mark
i've been set the question:

find the coordinates of the point on the y-axis which is equidistant from points A and D. point A is (0, 4) and D is (10, 6). i'm not sure how to go about figuring that one out. if any one could show me a step by step way of doing that, it would be appreciated.

thanks, Mark
Since the point you are looking for is on the y-axis it must be (0, y) for some number y. The distance from (0, y) to (0, 4) is |y- 4|. The distance from (0, y) to (10, 6) is $\sqrt{10^2+ (y- 6)^2}$. Since the point is equidistant from (0, 4) and (10, 6) we must have $|y- 4|= \sqrt{100+ (y- 6)^2}$.

Square both sides to get $(y- 4)^2= 100+ (y- 6)^2$, That is the same as $y^2- 8y+ 16= 100+ y^2- 12y+ 36$. Now the " $y^2$" terms cancel and you have an easy linear equation to solve for y.

I must say that, while earboth's method is correct, his graph is way off. It appears to give about (0, 17) as the solution and that is far from correct.