A) \[\left( -\frac{1}{5},\frac{16}{5} \right)\]
B) \[\left( \frac{14}{5},\frac{1}{5} \right)\]
C) \[\left( \frac{1}{5},\frac{1}{5} \right)\]
D) \[\left( \frac{14}{5},\frac{14}{5} \right)\]
E) \[(5,14)\]
Correct Answer: A
Solution :
Let\[A(5,-2),B(-1,2)\]and C (1, 4) are vertices of a triangle ABC. Slope of line \[BC=\frac{4-2}{1+1}=1.\] \[\therefore \]slope of line perpendicular to \[BC=-1\] So equation of altitude through A is \[(y+2)=-1(x-5)\] \[\Rightarrow \] \[x+y-3=0\] ...(i) and slope of \[CA=\frac{4+2}{1-5}=-\frac{3}{2}\] \[\therefore \]slope of a line perpendicular to\[CA=\frac{2}{3}\] so equation of altitude through C is \[(y-4)=\frac{2}{3}(x-1)\] \[\Rightarrow \] \[2y-12=2x-2\] \[\Rightarrow \] \[2x-3y=-10\] ...(ii) On solving Eqs. (i) and (ii), we get \[x=-\frac{1}{5},y=\frac{16}{5}\] \[\therefore \]Orthocentre of triangle is\[\left( -\frac{1}{5},\frac{16}{5} \right)\].
You need to login to perform this action.
You will be redirected in
3 sec
