Question

Orthocenter of the triangle whose vertices are (0,0) (2,-1) and (1,3) is -

• A. (-4,-1)
• B. $(-\frac{4}{7},-\frac{1}{7})$
• C. $(\frac{4}{7},\frac{1}{7})$
• D. (4,1)

Answer 1

The correct option is B $(-\frac{4}{7},-\frac{1}{7})$

Orthocenter is point of intersection of altitudes

$m_{AD}\cdot m_{BC}=-1$

$m_{BC}=\frac{3+1}{1-2}=-4$

$\Rightarrow m_{AD}=\frac{1}{4}$

$\therefore$ Equation of AD is $y=\frac{1}{4}x$

$m_{BE}\cdot m_{AC}=-1$, $m_{AC}=\frac{3}{1}$ $\Rightarrow m_{BE}=\frac{-1}{3}$

Equation of BE is $\frac{-1}{3}(x-2)=(y+1)$

$\Rightarrow -x+2=3y+3$

$\Rightarrow x+3y+1=0$

Point of intersection of AD & BE is 'H'(orthocentre)

$\Rightarrow y=\frac{x}{4}$ & $x+3y+1=0$, $\Rightarrow x=4y$

$4y+3y+1=0$ $\Rightarrow y=\frac{-1}{7}$

$\Rightarrow x=\frac{-4}{7}$

Orthocentre$(\frac{-4}{7},\frac{-1}{7})$.