Let the three sides of a triangle be on the lines $4x - 7y + 10 = 0$,$x + y = 5$,and $7x + 4y = 15$. Then the distance of its orthocentre from the orthocentre of the triangle formed by the lines $x = 0$,$y = 0$,and $x + y = 1$ is

  • A
    $5$
  • B
    $\sqrt{5}$
  • C
    $\sqrt{20}$
  • D
    $20$

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