JEE MainMathematicsStraight Lines
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
- A1
- B9
- C14
- 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 +