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  • Find the equations of the lines through the point of intersection of the lines x – y + 1 = 0 and 2x – 3y + 5 = 0 and whose distance from the point (3, 2) is 7/5.

Find the equations of the lines through the point of intersection of the lines x – y + 1 = 0 and 2x – 3y + 5 = 0 and whose distance from the point (3, 2) is 7/5.

Answers (1)

Given two lines are:   x-y+1=0 and 2x-3y+5=0  

 Solving these two equations gives us points of intersection  we get y=3 and x=2  

(x,y)=(2,3) 

 Let m be the slope of the required line  

 Then, equation of the line is y-3=mx-2 

 y-3=mx-2m 

 mx-y-2m+3=0….(1)   

 Since, the perpendicular distance from the point 3,2to the line is75then

d=\frac{\left | m*\left ( 3 \right ) -2 +3-2m \right |}{\sqrt{m^{2}+1^{2}}}

  \frac{7}{5}=\frac{\left | 3m+1-2m \right |}{\sqrt{m^{2}+1^{1}}}=\frac{m+1}{\sqrt{m^{2}+1^{2}}}

   Squaring both the sides, we get \frac{49}{25}=\frac{\left ( m+1 \right )^{2}}{m^{2}+1}

  49m2+1=25(m+1)2    

49m2+49=25m2+25+50m 

 25m2+25+50m-49m2-49=0  

 -24m2+50m-24=0  

 -12m2+25m-12=0  

Factorising, we get (3m-4)(4m-3)=0 

 3m-4=0 or 4m-3=0

  3m=4 or 4m=3  

 m=4/3 or 3/4 

  Putting the value of m=4/3 in equation 1 we get 4x/3-y-2(4/3)+3=0

  4x/3-y-8/3+3=0  

 4x/3-y=-1/3  

 4x-3y+1=0 

 Putting the value of m=3/4 in the equation we get 3x/4-y-2(3/4)+3=0 

3x/4-y-3/2+3=0 

3x/4-y+3/2=0  

 3x-4y+6=0 

  Hence, the required equation are 4x-3y+1=0  and  3x-4y+6=0   

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