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

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The correct Answer is:
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Since the given point `P(3,2)` and line is `x-2y=3`.
Slope of this line is `m_1(1)/(2)`
Let the slope of the required line is m.
then, `tan theta =|(m-(1)/(2))/(1+(1)/(2)m)|`
`rArr 1=pm((m-(1)/(2))/(1+(m)/(2)))[because tan 45^@=1]`.....(i)
Taking positive sign, `1+(m)/(2)=m-(1)/(2)`
`rArr m+(m)/(2)=1+(1)/(2)`
`rArr(m)/(2)+(3)/(2)rArrm=3`
Taking negative sign,
`1=-((m-(1)/(2))/(1+(m)/(2)))`
`rArr1+(m)/(2)=-m+(1)/(2)`
`rArrm+(m)/(2)=(1)/(2)-1`
`(3m)/(2)=(-1)/(2)rArrm=(-1)/(3)`
`therefore` First equation of the line is `y-2=3(x-3)`
`rArr3y-y-7=0`
and second equation of the line is
`y-2=-(1)/(3)(x-3)`
`rArr3y-6=-x+3`
`rArr x+3y-9=0`
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