Question

Two sides of a triangle are $y=m_{1}x$ and $y=m_{2}x.m_1,m_2$ are the roots of the equation $x^2+ax-1=0$. For all values of $a$, the orthocentre of the triangle lies at

• A. $(1,1)$
• B. $(2,2)$
• C. $(\frac{3}{2},\frac{3}{2})$
• D. $(0,0)$

Answer 1

Answer: (D)

$(0,0)$

$y=m_{1}x$ and $y=m_{2}x$ are two sides of triangle

$x^2+ax-1=0$ has roots $m_1,m_2$

$\therefore m_1{m}_2=-1$

i.e. $y=m_{1}x$ & $y=m_{2}x$ lines are perpendicular to each other

So, orthocentre of right angle triangle lies at verticles of ${90}^{\circ }$ angle

So, orthocentre of given triangle is $(0,0)$