find the point on the X- Axis which is equidistant from (2,- 5) and (-2,9)
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P(X,0) , A(2,-5) , B(-2,9)
PA = X1 = X, X2 =2 ,Y1 = 0 , Y2 = -5
PB = X1 = X, X2 = -2 ,Y1 = 0 , Y2 = 9
PA = PB. [ equal distance from A and B]
PA² = PB²
Using distance formula,
PA² = PB²
(X2-X1)²+(Y2-Y1)² = (X2-X1²+ (Y2-Y2)²
PUT VALUES AND SOLVE THIS.U WILL GET THE ANSWER
PA = X1 = X, X2 =2 ,Y1 = 0 , Y2 = -5
PB = X1 = X, X2 = -2 ,Y1 = 0 , Y2 = 9
PA = PB. [ equal distance from A and B]
PA² = PB²
Using distance formula,
PA² = PB²
(X2-X1)²+(Y2-Y1)² = (X2-X1²+ (Y2-Y2)²
PUT VALUES AND SOLVE THIS.U WILL GET THE ANSWER
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