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
⇒ √x2+49−14x+36 = √x2+9−6x+16
⇒ x2 - 14x + 85 = x2 - 6x + 25 ⇒ - 14x + 6x = 25 - 85 ⇒ 8x = 60
⇒ x =
15
2

∴ The required point is (
15
2
,0
)

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