Question

A triangle ABC with vertices $A\equiv (2,5),B\equiv (5,2)andC\equiv (-1,-1)$ inscribed on the curve $xy-x-y-3=0$ and $H\equiv (\alpha ,\alpha )$ is orthocentre. Let D, E and F be the feet of perpendiculars from A, B and C on BC, CA and AB respectively and inradius of $\Delta DEF$ is $r$ then

• A. orthocentre of triangle ABC is $(-1,-1)$
• B. equation of DE is $x+y-$ $\frac{13}{5}=0$
• C. $r=\frac{4}{5}\sqrt{2}$ units
• D. $r=\frac{2}{5}\sqrt{2}$ units

Answer 1

The correct option is D $r=\frac{2}{5}\sqrt{2}$ units

Slope of $BC=\frac{2+1}{5+1}=\frac{3}{6}=\frac{1}{2}$

Equation of $BC:\frac{y-2}{x-5}=\frac{1}{2}$

$\Rightarrow 2y-4=x-5$

$\Rightarrow x-2y=1$ ....$\left(1\right)$

Now, slope of $AD{\perp }^{r}BC=-2$

$AD$ passes through $A(2,5)$. Therefore equation of $AD:$

$\frac{y-5}{x-2}=-2$

$\Rightarrow y-5=4-2x$

$\Rightarrow 2x+y=9$ ....$\left(2\right)$

Solve $\left(1\right)$ and $\left(2\right)$ for $D$.

Multiply $\left(2\right)$ by $2$, $4x+2y=18$ ....$\left(3\right)$

Add $\left(1\right)$ and $\left(3\right)$,

$5x=19\Rightarrow x=\frac{19}{5}$

$y=9-2x=9-2\frac{19}{5}=9-\frac{38}{5}=\frac{7}{5}$

$\therefore D\equiv (\frac{19}{5},\frac{7}{5})$

Similarly, slope of $AC=\frac{5+1}{2+1}=2$

Equation of $AC:\frac{y-5}{x-2}=2$

$\Rightarrow y-5=2x-4$

$\Rightarrow y-2x=1$ ....$\left(4\right)$

Slope of $BE{\perp }^r=-\frac{1}{2}$

Equation of $BE:\frac{y-2}{x-5}=-\frac{1}{2}$

$\Rightarrow 2y-4=5-x$

$\Rightarrow x+2y=9$ ....$\left(5\right)$

Multiply $\left(5\right)$ by $2$,

$2x+4y=18$ ...$\left(6\right)$

Adding $\left(4\right)$ and $\left(6\right)$, we get

$5y=19\Rightarrow y=\frac{19}{5}$

$x=9-2\frac{19}{5}=\frac{7}{5}$

Therefore $E\equiv (\frac{7}{5},\frac{19}{5})$

Slope of $DE:\frac{\frac{19}{5}-\frac{7}{5}}{\frac{7}{5}-\frac{19}{5}}=-1$

Equation of $DE:\frac{y-\frac{19}{5}}{x-\frac{7}{5}}=-1$

$\Rightarrow y-\frac{19}{5}=\frac{7}{5}-x$

$\Rightarrow x+y=\frac{26}{5}$

Since the orthocentre $H(\alpha ,\alpha )$ lies on $AD$

$\Rightarrow 2\alpha +\alpha =9$

$\Rightarrow 3\alpha =9$

$\Rightarrow \alpha =3$

Therefore $H\equiv (3,3)$

Now, the inradius of $△DEF$ will be the perpendicular distance of $H$ from $DE$.

$\therefore r=\frac{3+3-\frac{26}{5}}{\sqrt{1+1}}=\frac{4}{5\sqrt{2}}$

$\Rightarrow r=\frac{2}{5}\sqrt{2}$

Hence, option $D$ is correct.