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Find the equation of the lines through the point (3, 2) which make an angle of 45° with the line x –2y = 3.

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प्रश्न

Find the equation of the lines through the point (3, 2) which make an angle of 45° with the line x –2y = 3.

योग
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उत्तर

Let the equation of line AB be: x – 2y = 3

or y = `1/2 "x" - 3`  

Then the slope of line AB = `1/2`

Let the PA line make an angle of 45° with it and its slope = m.

`± tan 45= ("m" - 1/2)/(1 + "m" xx 1/2)`

or `± 1 = (2"m" - 1)/("m" + 2)`

Taking +ve sign, 1 = `(2"m" - 1)/("m" + 2)`

or 2m – 1 = m + 2

∴ m = 3

2m –1

Taking – ve sign, –1 = `(2"m" - 1)/("m" + 2)`

or 2m – 1 = –m – 2

∴ 3m = –1

or m = `(-1)/3`

Hence, the equation of line PA is where point P = (3, 2) and m = `(-1)/3`.

y – 2 = `- (-1)/3 ("x" - 3)`

3y – 6 = – x + 3

or x + 3y – 9 = 0

Now when the slope is m = 3, then the equation of the line from the point P(3, 2),

y – 2 = 3(x – 3)

y – 2 = 3x – 9

or 3x – y – 7 = 0

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अध्याय 9: Straight Lines - Miscellaneous Exercise [पृष्ठ १७३]

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एनसीईआरटी Mathematics [English] Class 11
अध्याय 9 Straight Lines
Miscellaneous Exercise | Q 10. | पृष्ठ १७३

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