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  • A point equidistant from the lines 4x + 3y + 10 = 0, 5x – 12y + 26 = 0 and 7x + 24y – 50 = 0 is A. (1, –1) B. (1, 1) C. (0, 0) D. (0, 1)

A point equidistant from the lines 4x + 3y + 10 = 0, 5x – 12y + 26 = 0 and 7x + 24y – 50 = 0 is
A. (1, –1)
B. (1, 1)
C. (0, 0)
D. (0, 1)

Answers (1)

Given equations are 4x+3+10=0 …..(i)  

  5x-12y+26=0…......(ii) 

 and 7x+24y-50=0  

  Let p,qbe the point which is equidistant from the givenlines

  Now, we find the distance of (p,q) from the given lines  

Distance of point (x1,y1) from the equation Ax+By+C=0 

  d=\frac{\left | Ax+By+C \right |}{\sqrt{A^{2}+B^{2}}}

    Distance of (p,q) from equation i is d=\frac{\left | 4p+4q+10 \right |}{\sqrt{4^{2}+3^{2}}}=\frac{\left | 4p+4q+10 \right |}{5}

   Distance of (p,q) from equation ii  is d=\frac{\left | 5p-12q+26 \right |}{\sqrt{5^{2}+\left (-12 \right )^{2}}}=\frac{\left | 5p-12q+26 \right |}{13}

   Distance of (p,q) from equation iii  is d=\frac{\left | 7p-24q+50 \right |}{\sqrt{7^{2}+24^{2}}}=\frac{\left | 7p-24q+50\right |}{25}

  Given that (p,q) is equidistant from the given lines \frac{\left | 4p-3q+10 \right |}{5}=\frac{\left | 5p-12q+26 \right |}{13}=\frac{\left | 7p-24q+50\right |}{25}

  On putting the value of (p,q) as (0,0) we get \left | \frac{10}{5} \right |=\left | \frac{26}{13} \right |=\left | -\frac{50}{25} \right |

Hence, the correct option is c

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