मराठी

Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).

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प्रश्न

Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).

बेरीज
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उत्तर

Let (a, 0) be the point on the x axis that is equidistant from the points (7, 6) and (3, 4).

Accordingly, `sqrt((7 - a)^2 + (6 - 0)^2) = sqrt((3 - a)^2 + (4 - 0)^2)`

= `sqrt(49 + a^2 - 14a + 36) = sqrt(9 + a^2 - 6a + 16)`

= `sqrt(a^2 - 14a + 85) = sqrt(a^2 - 6a + 25)`

On squaring both sides, we obtain

a2 - 14a + 85 = a2 - 6a + 25

= -14a + 6a = 25 - 85

= -8a = -60

= `a = 60/8 = 15/2`

Thus, the required point on the x-axis is `(15/2, 0)`.

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पाठ 9: Straight Lines - EXERCISE 9.1 [पृष्ठ १५९]

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एनसीईआरटी Mathematics [English] Class 11
पाठ 9 Straight Lines
EXERCISE 9.1 | Q 4. | पृष्ठ १५९

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