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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.
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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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