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Find the point on the X-axis which is equidistant from (-3,4) and B(1,-4). |
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Answer» A(-3,4) and B(1,-4) Let P be the point on X-axis equidistant from points A and B . ` therefore PA = PB ` ….(1) P lies on X-axis ` therefore ` its y coordinate is 0. Let P(x,0) Now , PA = PB …[From (1)] By distance formula , ` ([x-(-3)]^(3) + (0-4)^(2))=sqrt((x-1)^(2) + [0-(-4)]^(2))` ` therefore sqrt((x +3)^(2) + (-4)^(2))= sqrt((x-1)^(2) + (4)^(2))` Squaring both the sides , `(x + 3)^(2) + 16 = (x-1)^(2) + 16` ` therefore x^(2) + 6x + 9 = x^(2) - 2x + 1` ` therefore 6x + 2x= 1 - 9 ` ` therefore 8x = - 8` ` therefore x = - (8)/(8) therefore x = - 1` ` therefore P(-1,0) ` The coordinates of point on X-axis equidistant from points A and B are (-1,0) . |
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