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प्रश्न
Find the point on the x-axis which is equidistant from (2, -5) and (-2, 9).
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उत्तर १
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 (x, 0) and (2, -5) = `sqrt((x-2)^2+(0-(-5))^2)`
= `sqrt((x-2)^2+(5)^2)`
Distance between (x, 0) and (-2, -9) = `sqrt((x-(-2))^2+(0-(-9))^2)`
= `sqrt((x+2)^2+(9)^2)`
By the given condition, these distances are equal in measure.
`sqrt((x-2)^2 +(5)^2)`
= `sqrt((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).
उत्तर २
Let (x, 0) be the point on the x axis. Then as per the question, we have
⇒ `sqrt((x-2)^2 +(0+5)^2)`
⇒ `sqrt((x+2)^2 + (0-9)^2)`
⇒ `sqrt((x-2)^2 +(5)^2)=sqrt((x+2)^2 + (9)^2)`
⇒ (x - 2)2 + (5)2 = (x + 2)2 + (-9)2 ...(Squaring both sides)
⇒ x2 - 4x + 4 + 25 = x2 + 4x + 4 + 81
8x = 25 - 81
8x = -56
x = -7
Therefore, the point is (−7, 0).
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