question_answer

The orthocentre of the triangle with vertices\[O(0,\,\,0),\,\,A\left( 0,\,\,\frac{3}{2} \right),\,\,B(-5,\,\,0)\]is

A) \[\left( \frac{5}{2},\,\,\frac{3}{4} \right)\]                            

B) \[\left( \frac{-5}{2},\,\,\frac{3}{4} \right)\]

C) \[\left( -5,\,\,\frac{3}{2} \right)\]             

D)        \[(0,\,\,0)\]

Correct Answer: D

Solution :

\[\Delta AOB\]is the given triangle. Slope of\[AB=\frac{\frac{3}{2}-9}{0+5}=\frac{3}{10}\] Slope of\[BO=\frac{0-0}{0+5}=0\] The equation of line passing through\[A\]and perpendicular to\[BO\] is\[y-0=0\left( x-\frac{3}{2} \right)\] \[\Rightarrow \]                               \[y=0\]                                 ... (i) and equation of line passing through 0 and perpendicular to\[AB\]is\[y-0=\frac{10}{3}(x-0)\] \[\Rightarrow \]                               \[y=-\frac{10}{3}x\]                        ? (ii) The intersection point of Eqs. (i) and (ii) is (0, 0), which is the required orthocentre.