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Find the equation of a line with slope 3...

Find the equation of a line with slope 3 and the length of the perpendicular from the origin equal to `sqrt(10)`

Text Solution

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Let c be the intercept on y-axis
Then, the equation of the line is
`y=3x+c`
`implies-33x+y=c`
`implies(-3)/(sqrt((-3)^(2)+1^(2)))x+(y)/(sqrt((-3)^(2)+1^(2)))=(c)/(sqrt((-3)^(2)+1^(2)))`
`implies(-3)/(sqrt10)x+(y)/(sqrt10)=(c)/(sqrt10)`
This is the normal form of the equation
Thus, `(c)/(sqrt10)` denotes the length of perpendicular from the origin. But, the length of the perpendicular from the origin is `sqrt10`
`therefore` The required lines are `-3x+y=+-10`
or `y=3x+-10`
`therefore|(c)/(sqrt10)|=sqrt10`
`implies(|c|)/(sqrt10)=sqrt10`
`implies|c|=1=+-10.`
`therefore` The required lies are `-3x+y=+-10or y=3x+-10`
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