document.write( "Question 624446: If (-1, 2) is equidistant with (4,4)and (x, 4) find x \n" ); document.write( "
Algebra.Com's Answer #392754 by Theo(13342)\"\" \"About 
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(-1,2) is equidistant with (4,4) and (x,4)
\n" ); document.write( "the y value has to be 4 which is on the line y = 4
\n" ); document.write( "the x value can be any value as long as the y value is 4.
\n" ); document.write( "there is only one place where it can be, however.
\n" ); document.write( "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).
\n" ); document.write( "if the distance is the same, they are equidistant.
\n" ); document.write( "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).
\n" ); document.write( "the distance between the point (-1,2) and the point (4,4) is found using the following formula:
\n" ); document.write( "d = sqrt((4+1)^2 + (4-2)^2) which is simplified to:
\n" ); document.write( "d = sqrt(5^2 + 2^2) which is further simplified to:
\n" ); document.write( "d = sqrt(29)
\n" ); document.write( "that has to be the distance between the point (-1,2) and the new point of (x,4)
\n" ); document.write( "using the same formula, we get:
\n" ); document.write( "d = sqrt((x+1)^2 + (4-2)^2) which is simplified to:
\n" ); document.write( "d = sqrt((x+1)^2 + 2^2) which is further simplified to:
\n" ); document.write( "d = sqrt((x+1)^2 + 4)
\n" ); document.write( "since d is equal to sqrt(29), this formula becomes:
\n" ); document.write( "sqrt(29) = sqrt((x+1)^2 + 4)
\n" ); document.write( "if we square both sides of this equation, we get:
\n" ); document.write( "29 = (x+1)^2 = 4
\n" ); document.write( "subtract 4 from both sides of this equation to get:
\n" ); document.write( "(x+1)^2 = 25
\n" ); document.write( "take the square root of both sides of this equation to get:
\n" ); document.write( "x+1 = +/- 5
\n" ); document.write( "when x+1 = 5, we get x = 4
\n" ); document.write( "when x+1 = -5, we get x = -6
\n" ); document.write( "the answer is that x can be either 4 or -6.
\n" ); document.write( "since we already have the point (4,4), we then assume the other point is equal to (-6,4)
\n" ); document.write( "to confirm, we calculate the distance between the point (-1,2) and the point (-6,4).
\n" ); document.write( "that distance is calculated as:
\n" ); document.write( "d = sqrt((-6-(-1))^2 + (4-2)^2) which simplifies to:
\n" ); document.write( "d = sqrt((-6+1)^2 + (2)^2 which further simplifies to:
\n" ); document.write( "d = sqrt((-5)^2 + 4) which further simplifies to:
\n" ); document.write( "d = sqrt(25+4) which further simplifies to:
\n" ); document.write( "d = sqrt(29).
\n" ); document.write( "that's the same distance from the point (4,4) so we're good.
\n" ); document.write( "the new point is (-6,4).
\n" ); document.write( "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.
\n" ); document.write( "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.
\n" ); document.write( "what you have is 2 right triangles with the same leg lengths which makes their hypotenuse the same.
\n" ); document.write( "the hypotenuse is equal to square root of (leg1^2 + leg2^2)
\n" ); document.write( "that's the formula of length of line segment equals sqrt ((x1-x2)^2 + (y1-y2)^2
\n" ); document.write( "a reference is shown below:
\n" ); document.write( "http://www.regentsprep.org/regents/math/geometry/GCG3/Ldistance.htm\r
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