NCERT Class XI Mathematics - Straight Lines - Solutions
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Question : 4
Total: 74
Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).
Solution:
Let the point be P(x, y). Since it lies on the x-axis
∴ y = 0 i.e., required point be (x, 0). Since the required point is equidistant from points A(7, 6) and B(3, 4) ⇒ PA = PB
⇒√ ( x − 7 ) 2 + ( 0 − 6 ) 2 = √ ( x − 3 ) 2 + ( 0 − 4 ) 2
⇒√ x 2 + 49 − 14 x + 36 = √ x 2 + 9 − 6 x + 16
⇒x 2 - 14x + 85 = x 2 - 6x + 25 ⇒ - 14x + 6x = 25 - 85 ⇒ 8x = 60
⇒ x =
∴ The required point is(
, 0 )
∴ y = 0 i.e., required point be (x, 0). Since the required point is equidistant from points A(7, 6) and B(3, 4) ⇒ PA = PB
⇒
⇒
⇒
⇒ x =
∴ The required point is
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