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Question 624446: If (-1, 2) is equidistant with (4,4)and (x, 4) find x
Found 2 solutions by ewatrrr, Theo:
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website!
Hi,
If (-1, 2) is equidistant with (4,4)and (x, 4) find x
(-1, 2)
(4, 4) 
(-1, 2)
( x, 4) 
(1 + x)^2 = 25, x +1 = ±5, x = -6 or x = 4
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! (-1,2) is equidistant with (4,4) and (x,4)
the y value has to be 4 which is on the line y = 4
the x value can be any value as long as the y value is 4.
there is only one place where it can be, however.
that place is when the distance between the point (-1,2) and the new value of (x,4) is equal to the distance between the point (-1,2) and (4,4).
if the distance is the same, they are equidistant.
so, the problem is find the new point on (x,4) that has the same distance between it and the point (-1,2) as the old distance between the point (4,4) and the point (-1,2).
the distance between the point (-1,2) and the point (4,4) is found using the following formula:
d = sqrt((4+1)^2 + (4-2)^2) which is simplified to:
d = sqrt(5^2 + 2^2) which is further simplified to:
d = sqrt(29)
that has to be the distance between the point (-1,2) and the new point of (x,4)
using the same formula, we get:
d = sqrt((x+1)^2 + (4-2)^2) which is simplified to:
d = sqrt((x+1)^2 + 2^2) which is further simplified to:
d = sqrt((x+1)^2 + 4)
since d is equal to sqrt(29), this formula becomes:
sqrt(29) = sqrt((x+1)^2 + 4)
if we square both sides of this equation, we get:
29 = (x+1)^2 = 4
subtract 4 from both sides of this equation to get:
(x+1)^2 = 25
take the square root of both sides of this equation to get:
x+1 = +/- 5
when x+1 = 5, we get x = 4
when x+1 = -5, we get x = -6
the answer is that x can be either 4 or -6.
since we already have the point (4,4), we then assume the other point is equal to (-6,4)
to confirm, we calculate the distance between the point (-1,2) and the point (-6,4).
that distance is calculated as:
d = sqrt((-6-(-1))^2 + (4-2)^2) which simplifies to:
d = sqrt((-6+1)^2 + (2)^2 which further simplifies to:
d = sqrt((-5)^2 + 4) which further simplifies to:
d = sqrt(25+4) which further simplifies to:
d = sqrt(29).
that's the same distance from the point (4,4) so we're good.
the new point is (-6,4).
if you create a right triangle from the point (-1,2) to the point (4,4), you will see that the absolute value of the difference in the x values and the y values is 5 for x and 2 for y.
if you create a right triangle from the point (-1,2) to the point (-6,4), you will see that the absolute value of the difference in the x values and the y values is 5 for x and 2 for y.
what you have is 2 right triangles with the same leg lengths which makes their hypotenuse the same.
the hypotenuse is equal to square root of (leg1^2 + leg2^2)
that's the formula of length of line segment equals sqrt ((x1-x2)^2 + (y1-y2)^2
a reference is shown below:
http://www.regentsprep.org/regents/math/geometry/GCG3/Ldistance.htm
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