question_answer

The orthocenter of a triangle formed by the lines \[x+y=1,\,\,2x+3y=6\]and \[4x-y+4=0\] lies in the

A)  Ist  quadrant   

B)  IInd  quadrant

C)  IIIrd quadrant  

D)  IVth quadrant

Correct Answer: A

Solution :

The equation of lines are \[x+y=1,2x+3y=6\] and \[4x-y+4=0\] On solving equation Ist and IInd, and Ist and IIIrd, we get \[A(-3,\,4)\] and \[B\left( -\frac{3}{5},\,\frac{8}{5} \right).\] The equation of perpendicular line of the line \[4x-y+4=0\] and passes through a point \[A(-3,4)\] is \[3x-2y+5=0\] ?..(i) Also,  the equation of perpendicular line to the line \[2x+3y=6\] and passes through a point \[B\left( -\frac{3}{5},\frac{8}{5} \right)\] is \[3x-2y+5=0\] ?.. (ii) On solving Eqs. (i) and (ii), we get the orthocenter \[\left( \frac{3}{7},\frac{22}{7} \right)\] which is lies in 1st quadrant.