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Find the equation of the line which satisfy the given condition: Passing though (2,23) and is inclined with the x-axis at an angle of 75°. - Mathematics

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प्रश्न

Find the equation of the line which satisfy the given condition:

Passing though `(2, 2sqrt3)` and is inclined with the x-axis at an angle of 75°.

योग

उत्तर

Since, the line is inclined at 75° to the x-axis, then the slope of the line is

m = tan 75° = tan (45° + 30°)

= `("tan" 45° + "tan" 30°)/(1 - "tan" 45° "tan" 30°)`

= `(1 + 1/sqrt3)/(1 - 1. 1/sqrt3)`

= `((sqrt3 + 1)/sqrt3)/((sqrt3 - 1)/sqrt3)`

= `(sqrt3 + 1)/(sqrt3 - 1)`

We know that the equation of the line m passing through the point (x0, y0) whose slope is (y − y0) = m(x − x0).

If a line passes through an angle of 75° with the x-axis and the equation of the line is given as.

`("y" - 2sqrt3) = (sqrt3 + 1)/(sqrt3  -1) ("x"- 2)`

`("y" - 2sqrt3) (sqrt3 - 1) = (sqrt3 + 1) ("x" - 2)`

`"y"(sqrt3 - 1) - 2sqrt3(sqrt3 - 1) = "x"(sqrt3 + 1) - 2(sqrt3 + 1)`

`(sqrt3  + 1) "x" - (sqrt3 - 1) "y" = 2sqrt3 + 2 - 6 + 2sqrt3`

`(sqrt3  + 1) "x" - (sqrt3 - 1) "y" = 4sqrt3 - 4`

∴ `(sqrt3  + 1) "x" - (sqrt3 - 1) "y" = 4(sqrt3 - 1)`

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अध्याय 10: Straight Lines - Exercise 10.2 [पृष्ठ २१९]

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एनसीईआरटी Mathematics [English] Class 11
अध्याय 10 Straight Lines
Exercise 10.2 | Q 4 | पृष्ठ २१९

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