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Question 7
Find the point on the x-axis which is equidistant from (2, - 5) and (- 2, 9).
Question 7
Find the point on the x-axis which is equidistant from (2, - 5) and (- 2, 9).
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Solution
We have to find a point on x-axis. Therefore, its y-coordinate will be 0.
Let the point on x-axis be (x,0)
Distance between the points is given by
√[(x1−x2)2+(y1−y2)2]
Distance between (x,0) and (2,−5)
=√[(x−2)2+(0−(−5))2]=√[(x−2)2+(5)2]
Distance between (x,0) and (−2,9)
=√[(x−(−2))2+(0−(9))2]=√[(x+2)2+(9)2]
By the given condition,
=√[(x−2)2+(5)2]=√[(x+2)2+(9)2]
=(x−2)2+25=(x+2)2+81
=x2+4−4x+25=x2+4+4x+81
=8x=25−81
=8x=−56
=x=−7
Therefore, the point is (−7,0).
We have to find a point on x-axis. Therefore, its y-coordinate will be 0.
Let the point on x-axis be (x,0)
Distance between the points is given by
√[(x1−x2)2+(y1−y2)2]
Distance between (x,0) and (2,−5)
=√[(x−2)2+(0−(−5))2]=√[(x−2)2+(5)2]
Distance between (x,0) and (−2,9)
=√[(x−(−2))2+(0−(9))2]=√[(x+2)2+(9)2]
By the given condition,
=√[(x−2)2+(5)2]=√[(x+2)2+(9)2]
=(x−2)2+25=(x+2)2+81
=x2+4−4x+25=x2+4+4x+81
=8x=25−81
=8x=−56
=x=−7
Therefore, the point is (−7,0).
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