what point on x axis is equidistant from the points A(7,6) and B(-3,4) ?
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Answered by
3
let p be the point which is equidistant frm da pnts given
p(k,0)
Ap = Bp
√(k-7)^2+(6)^2 =√(k+3)^2 + 4^2
k=3
p(k,0)
Ap = Bp
√(k-7)^2+(6)^2 =√(k+3)^2 + 4^2
k=3
Answered by
10
Any point on x-axis is represented by P (x, 0). where x is a real number.
Distance AP = √[(7-x)²+(6-0)²]
Distance BP = √[(-3-x)²+(4-0)²]
if they are equal, then : 49+x²-14x + 36 = 9+x²+6x+16
20 x = 60
=> x = 3
(3, 0) is the point.
Distance AP = √[(7-x)²+(6-0)²]
Distance BP = √[(-3-x)²+(4-0)²]
if they are equal, then : 49+x²-14x + 36 = 9+x²+6x+16
20 x = 60
=> x = 3
(3, 0) is the point.
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