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  • State whether the statements are true or false. The equation of the line joining the point (3, 5) to the point of intersection of the lines 4x + y – 1 = 0 and 7x – 3y – 35 = 0 is equidistant from the points (0, 0) and (8, 34).

State whether the statements are true or false.

The equation of the line joining the point (3, 5) to the point of intersection of the lines 4x + y – 1 = 0 and 7x – 3y – 35 = 0 is equidistant from the points (0, 0) and (8, 34).

Answers (1)

Given two lines are 4x+y-1=0…(i ) and   7x-3y-35=0…..(ii)  

 Now, point of intersection of these lines can be find out as x=2 and y=-7  

 To find the equation of the line joining the point (3,5) and (2,-7) 

y-y_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\left ( x-x_{1} \right )

y-5=\frac{-7-5}{2-3}\left ( x-3 \right )

 y-=-\frac{12}{-1}\left ( x-3 \right )

  y-5=12x-3   

  y-5=12x-36 

 12x-y-31=0……iv 

  Now, the distance of equation (iv) from the point (0,0) is d=\frac{\left |Ax+By+C \right |}{\sqrt{A^{2}+B^{2}}}

d=\frac{\left |12(8)-34-31 \right |}{\sqrt{12^{2}+(-1)}}

 

  d=\frac{\left | 31\right |}{\sqrt {145}}

   Now the distance of

equation iv from the point (8,34) is \frac{\left |Ax+By+C \right |}{\sqrt{A^{2}+B^{2}}}

 d=\frac{\left |12(8)-34-31 \right |}{\sqrt{12^{2}+(-1)}}=\frac{\left | 31\right |}{\sqrt {145}}

   Hence, the equation of line 12x-y-31=0 is equidistant from (0,0)and (8,34)  

Hence, the given statement is True. 

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infoexpert22

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