Question 1045568: locate the point which satisfies the given conditions.
Equidistant from (3,8), (5,2) and (-3,-4)
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! The point has coordinates x,y
Distance from (3,8) using distance formula is sqrt[(x-3)^2+(y-8)^2]=[x^2-6x+9+y^2-16y+64]^(1/2)
Distance from (5,2) is sqrt[(x-5)^2+(y-2)^2]=[x^2-10x+25+y^2-4y+4]^(1/2)
Distance from (-3,-4) is sqrt ((x+3)^2+(y+4)^2)=[x^2+6x+9+y^2+8y+16}^(1/2)
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Set the first and third equal to each other, since they are both equidistant. Then square them, removing the (1/2) exponent.
[x^2-6x+9+y^2-16y+64]=[x^2+6x+9+y^2+8y+16}
x^2 and y^2 on both sides cancel
-6x-16y+73=6x+8y+25
put x and y on the same side, put the constant on the other side
48=12x+24y
divide by 12 to simplify it.
4=x+2y
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Now set the first and second equal to each other and square them.
[x^2-6x+9+y^2-16y+64]=[x^2-10x+25+y^2-4y+4] remove x^2 and y^2
-6x-16y+73=-10x-4y+29
44=-4x+12y
Divide by 4 both sides.
11=-x+3y
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4=x+2y
11=-x+3y
add them
15=5y
y=3
substitute
x=-2
check in the other equation
(-2,3) ANSWER
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Distance from (-2,3) to( 3,8), (5,2) and (-3,-4)
The first is sqrt (25+25), the second is sqrt (49+1) and the third is sqrt (1+49). They are identical.
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