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महाराष्ट्र राज्य शिक्षण मंडळएस.एस.सी (इंग्रजी माध्यम) इयत्ता १० वी

Find the point on the X–axis which is equidistant from A(–3, 4) and B(1, –4). - Geometry Mathematics 2

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प्रश्न

Find the point on the X–axis which is equidistant from A(–3, 4) and B(1, –4).

बेरीज

उत्तर

Let point C be on the X-axis which is equidistant from points A and B.

Point C lies on X-axis.

∴ Its y co-ordinate is 0.

Let C = (x, 0)

C is equidistant from points A and B.

∴ AC = BC

According to distance formula,

`∴ sqrt([x  –  (– 3)]^2 + (0  –  4)^2] = sqrt((x  –  1)^2 + [0  –  ( –  4)]^2)`

∴ [x – (– 3)]2 + (0 – 4)2 = (x – 1)2 + [0 – ( – 4)]2    ...(Squaring both the sides)

∴ (x + 3)2  + (– 4)2 = (x – 1)2 + 42

∴ x2 + 6x + 9 + 16 = x2 – 2x + 1 + 16

∴ `cancelx^2 + 6x + 25 = cancelx^2  –  2x + 17`

∴ 6x + 2x = 17 – 25

∴ 8x = – 8

∴ `"x" = (-8)/8`

∴ x = –1

∴ The point on the X-axis, which is equidistant from points A and B, is (–1, 0).

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पाठ 5: Co-ordinate Geometry - Practice Set 5.1 [पृष्ठ १०७]

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बालभारती Geometry (Mathematics 2) [English] 10 Standard SSC Maharashtra State Board
पाठ 5 Co-ordinate Geometry
Practice Set 5.1 | Q 3 | पृष्ठ १०७

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