Question

If the ends of the base of an isosceles triangle are at $(2,0)$ and $(0,1)$ and the equation of one side is $x=2$, then the orthocentre of the triangle is

• A. $(\frac{3}{4},\frac{3}{2})$
• B. $(\frac{5}{4},1)$
• C. $(\frac{3}{4},1)$
• D. $(\frac{4}{3},\frac{7}{12})$

Answer 1

The correct option is B $(\frac{5}{4},1)$

$AB=AC$ $⟹y^2=2^2+(y-1{)}^{}⟹2y=4+1⟹y=\frac{5}{2}$
now CF will be perpendicular to AB

Using the concept: line perpendicular to
$ax+by+c=0$ is $bx-ay+c^{\prime }=0$

so equation of CF: $y=constant$ since it passes through C $constant=1$

equation of BC$:\frac{x}{2}+\frac{y}{1}=1⟹x+2y=2$ so equation of AD : $2x-y=c$ A passes through it $\therefore c=4-\frac{5}{2}=\frac{3}{2}$

$CF:y=1$ $AD:2x-y=\frac{3}{2}$ intersection of these two will give orthocentre $\therefore H\equiv (\frac{5}{4},1)$