हिंदी

Find the joint equation of the line passing through the origin and having slopes 1 + 3 and 1 - 3

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प्रश्न

Find the joint equation of the line passing through the origin and having slopes 1 + `sqrt3` and 1 - `sqrt3`

योग
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उत्तर

Let l1 and l2 be the two lines. Slopes of l1 is 1 + `sqrt3` and that of l2 is 1 - `sqrt3`

Therefore the equation of a line (l1) passing through the origin and having slope is

y = `(1 + sqrt3)"x"`

∴ `(1 + sqrt3)"x" - "y" = 0`   ...(1)

Similarly, the equation of the line (l2) passing through the origin and having slope is

y = `(1 - sqrt3)"x"`

∴ `(1 - sqrt3)"x" - "y" = 0`   ...(2)

From (1) and (2) the required combined equation is

`[(1 + sqrt3)"x" - "y"][(1 - sqrt3)"x" - "y"] = 0`

∴ `(1 + sqrt3)"x"[(1 - sqrt3)"x" - "y"] - "y"[(1 - sqrt3)"x" - "y"] = 0` 

∴ `(1 - sqrt3)(1 + sqrt3)"x"^2 - (1 + sqrt3)"xy" - (1 - sqrt3)"xy" + "y"^2 = 0`

∴ `((1)^2 - (sqrt3)^2)"x"^2 - [(1 + sqrt3) + (1 - sqrt 3)]"xy" + "y"^2 = 0`

∴ `(1 - 3)"x"^2 - 2"xy" + "y"^2 = 0`

∴ `- 2"x"^2 - "2 xy" + "y"^2 = 0`

∴ 2x2 + 2xy - y2 = 0

This is the required combined equation.

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अध्याय 4: Pair of Straight Lines - Miscellaneous Exercise 4 [पृष्ठ १३१]

APPEARS IN

बालभारती Mathematics and Statistics 1 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 4 Pair of Straight Lines
Miscellaneous Exercise 4 | Q 1.08 | पृष्ठ १३१

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