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Equations of the lines through the point (3, 2) and making an angle of 45° with the line x – 2y = 3 are ____.

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Given equation is x-2y=3  

 x-3=2y 

y=\frac{1}{2}x+\left ( -\frac{3}{2} \right ).................(i)

The slope of the equation can be found by comparing with y=mx+b form  

So, m_{1}=\frac{1}{2}

 We have to find an equation which is passing through the point (3,2)   

A line passing through the point x1,y1 has an equation y-y1=m(x-x1

 So, here x1=3 and y1=2 

  y-2=m(x-3)....…(ii) 

  Now, it is given that the angle between the given two lines is 45

 \tan \theta =\left | \frac{\left ( m_{1}-m_{2} \right )}{1+m_{1}m_{2}} \right |

  Putting the values of m1and m2 in above equation we get tan450 = \left | \frac{ m-\frac{1}{2} }{1+m*\frac{1}{2}} \right |

1=\left | \frac{ 2m-1 }{2+m} \right |

1=\pm \frac{ 2m-1 }{2+m}

     2m-1=2+m   or-(2m-1)=2+m   

2m-m=2+1  or -2m+1-m=2     

 m=3 or-3m=1 or   m=-1/3 

   Putting the value of m=3 in equation (ii) we get y-2=3(x-3) 

   y-2=3x-9   

  3x-y-9+2=0   

  3x-y-7=0 

 Putting the value of m=-1/3 in equation (ii) we get  y-2= -1/3(x-3)  

 3(y-2)=3-x

   3y-6=3-x  

 x+3y-6-3=0 

  x+3y-9=0

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