SOLUTION: Find the point y-axis that is equidistant from (5,1) and (-3, -1)

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Question 1203326: Find the point y-axis that is equidistant from (5,1) and (-3, -1)
Found 2 solutions by math_tutor2020, josgarithmetic:
Answer by math_tutor2020(3816) About Me  (Show Source):
You can put this solution on YOUR website!

Answer: (0,4)

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Explanation


I'll go over two methods.

Method 1

The given points are (5,1) and (-3,-1) which I'll label A and B respectively.

The mystery point is of the form (0,y) residing on the y axis.
Let's call this point C.

A = (5,1)
B = (-3,-1)
C = (0,y)

The goal is to determine the value of y such that segment AC and segment BC have the same length.

Compute the distance from A to C
d+=+sqrt%28+%28x2-x1%29%5E2%2B%28y2-y1%29%5E2+%29

d+=+sqrt%28+%280-5%29%5E2%2B%28y-1%29%5E2+%29

d+=+sqrt%28+%28-5%29%5E2%2B%28y-1%29%5E2+%29

d+=+sqrt%28+25%2B%28y-1%29%5E2+%29

d+=+sqrt%28+25%2B%28y%5E2-2y%2B1%29+%29

d+=+sqrt%28+y%5E2-2y%2B26+%29

Then do the same for the distance from B to C
d+=+sqrt%28+%28x2-x1%29%5E2%2B%28y2-y1%29%5E2+%29

d+=+sqrt%28+%280-%28-3%29%29%5E2%2B%28y-%28-1%29%29%5E2+%29

d+=+sqrt%28+%280%2B3%29%5E2%2B%28y%2B1%29%5E2+%29

d+=+sqrt%28+%283%29%5E2%2B%28y%2B1%29%5E2+%29

d+=+sqrt%28+9%2B%28y%5E2%2B2y%2B1%29+%29

d+=+sqrt%28+y%5E2%2B2y%2B10+%29

Equate those expressions to allow us to solve for y.

distance A to C = distance B to C

sqrt%28+y%5E2-2y%2B26+%29+=+sqrt%28+y%5E2%2B2y%2B10+%29

y%5E2-2y%2B26+=+y%5E2%2B2y%2B10 Square both sides

-2y%2B26+=+2y%2B10 The y^2 terms cancel

-2y-2y+=+10-26

-4y+=+-16

y+=+-16%2F%28-4%29

y+=+4

We go from the template (0,y) to (0,4)
The point (0,4) is equidistant from A(5,1) and B(-3,-1) such that it is on the y axis.

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Method 2

The given points are
A = (5,1)
B = (-3,-1)

Compute the midpoint of segment AB.
We add up the coordinates and divide in half
x: (5+(-3))/2 = 1
y: (1+(-1))/2 = 0

The midpoint of segment AB is located at (1,0) which I'll call point D.

Now compute the slope of line AB
m = (y2-y1)/(x2-x1)
m = (-1-1)/(-3-5)
m = (-2)/(-8)
m = 1/4
Line AB has a slope of 1/4.
The perpendicular line has a slope of -4/1 = -4. Apply the negative reciprocal. We flip the fraction and flip the sign.

The goal now is to find the equation of the perpendicular line through the midpoint D(1,0)

In other words: we want the equation of the perpendicular bisector of segment AB.

Apply point-slope form
y - y1 = m(x - x1)
y - 0 = -4(x - 1)
y = -4x + 4
This equation has slope -4 and y-intercept 4

The y intercept of this equation is the exact point equidistant from A and B, and is on the y axis.
I'll leave the proof to the reader as to why this works. A hint would be to draw segments CA, CB, and CD.
Then prove triangle ACD is congruent to triangle BCD.

I recommend using graphing tools like GeoGebra or Desmos.

Answer by josgarithmetic(39616) About Me  (Show Source):
You can put this solution on YOUR website!
On the y-axis, some point (0,y).
That point equally distant from (5,1) and (-3, -1);
Use the distance formula.

sqrt%28%285-0%29%5E2%2B%281-y%29%5E2%29=sqrt%28%28-3-0%29%5E2%2B%28-1-y%29%5E2%29
Simplify and solve this.

25%2B%281-y%29%5E2=9%2B%28-1-y%29%5E2
16=%28-1-y%29%5E2-%281-y%29%5E2
16=1%2B2y%2By%5E2-%281-2y%2By%5E2%29
16=4y
highlight%28y=4%29