Find a point on x axis which is equidistant from( 6,3) and (3,0)
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Let the two points be A ( 6,3 ) , B ( 3,0 )
and the point on x axis is P ( x,0 )
They are equidistant from each other
AP = BP
by applying the distance formula
and the point on x axis is P ( x,0 )
They are equidistant from each other
AP = BP
by applying the distance formula
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A(6,3) (x1 y1)
B(3,0) (x2 y2)
P(x,0) (x y)
AP = BP
by distance formula,
AP = √[(x1-x)²+(y1-y)²]
= √[(6-x)²+(3-0)²]
= √[36+x²-12x+9]
= √[x²-12x+45]
BP = √[(x2-x)²+(y2-y)²]
= √[(3-x)²+(0-0)²]
= √[x²-6x+9]
as AP = BP
therefore
√[x²-12x+45] = √[x²-6x+9]
x²-12x+45 = x²-6x+9
12x-45 = 6x-9
4x-15 = 2x-3
2x = 12
x = 6
therefore the point is P(6,0)
B(3,0) (x2 y2)
P(x,0) (x y)
AP = BP
by distance formula,
AP = √[(x1-x)²+(y1-y)²]
= √[(6-x)²+(3-0)²]
= √[36+x²-12x+9]
= √[x²-12x+45]
BP = √[(x2-x)²+(y2-y)²]
= √[(3-x)²+(0-0)²]
= √[x²-6x+9]
as AP = BP
therefore
√[x²-12x+45] = √[x²-6x+9]
x²-12x+45 = x²-6x+9
12x-45 = 6x-9
4x-15 = 2x-3
2x = 12
x = 6
therefore the point is P(6,0)
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