# points on axis

• Aug 25th 2009, 02:28 AM
mark
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
• Aug 25th 2009, 02:36 AM
Prove It
Quote:

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...
• Aug 25th 2009, 02:39 AM
mark
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
• Aug 25th 2009, 03:55 AM
earboth
Quote:

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.
4. Equation of the line passing through the midpoint perpendicular to AD
5. Point of intersection of this line and the y-axis
• Aug 25th 2009, 04:20 AM
mark
thanks earboth, that was a good explanation
• Aug 25th 2009, 08:10 AM
HallsofIvy
Quote:

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.