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The equations of the sides AB and AC of a triangle ABC are 3x - 2y + 35 = 0 and 3x + 2y - 41 = 0 respectively. If the orthocentre of the triangle is H (1, 9) , then the perpendicular distance from the origin to the line containing the side BC is:

Options

  1. A1
  2. B9
  3. C14
  4. D25

Correct answer

A. 1

Step-by-step solution

Let the vertices of the triangle be A , B , and C . The slope of the side AC is given by - 3 2 . The altitude from vertex B passes through the orthocentre H (1, 9) and is perpendicular to AC . Therefore, the slope of the altitude BH is 2 3 . The equation of the altitude BH is: y - 9 = 2 3 (x - 1) 2x - 3y + 25 = 0 Vertex B is the intersection of the side AB and the altitude BH . Solving the equations 3x - 2y + 35 = 0 and 2x - 3y + 25 = 0 simultaneously: Multiplying the first by 3 and the second by 2 gives 9x - 6y +

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