Question

Find the equation of the line passing through the point (1,3)and making an angle of 45° with
the line x-3y+4=0​

Answer 1

Refer to attachment for diagram.

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${\huge{\underline{\sf {\pink{Answer}}}}}$

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$\boxed{ 2x \: + \: y \: - \: 5 \: = \: 0 }$

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${\huge{\underline{\sf {\pink{Explanation :}}}}}$

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Given :

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Line : x - 3y + 4 = 0

$\sf \theta = 45 \degree$

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To Find :

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Equation of required line.

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Solution :

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$\sf \: Slope \: of \: line \: = \frac{ - (coefficient \: of \: x)}{coefficient \: of \: y}$

= $\frac{-(-3)}{1}$

= 3

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Angle = 45°

$\sf \: tan \: \theta = \frac{{m}_{1} - {m}_{2}}{1 + {m}_{1}{m}_{2}}$

↬ $\sf \: tan \: 45° = \frac{{m}_{1} - 3}{1 + {m}_{1} 3}$

↬ $\sf \: 1 = \frac{m - 3}{1 + 3m}$

↬ 1 + 3m = m - 3

↬ 2m = - 4

↬ m = -2

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Given point on the line = ( 1 , 3 )

Slope of given line = - 2

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Equation of line :

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$\sf \: y - {y}_{1} = m(x - {x}_{1})$

⇢ $\sf \: y - 3 = -2(x - 1)$

⇢ $\sf \: y - 3 = -2x + 2$

⇢ $\sf \: 2x + y - 5 = 0$