$(a)\;(0,2)\qquad(b)\;(0,1)\qquad(c)\;(2,0)\qquad(d)\;(1,0)$

- Orthocentre is the interesting point of altitudes.
- Slope of the line joining $(x_1,y_1)$ and $x_2,y_2)$ is $\large\frac{y_2-y_1}{x_2-x_1}$

Orthocentre is the interesting point of altitudes.

Step 1

Find the equation of the altitude which is perpendicular to BC and passes through A.

Slope of $BC=\large\frac{0}{1+1}$$=0=m_1$

Since the altitude through $A$ is $\perp$ to $BC$,

slope of the altitude $ =-\large\frac{1}{0}$

Hence the equation of the altitude passing through $A(0,1)$ is

$(y-1)=\large\frac{-1}{0}$$(x-0)$

$i.e.,\:x=0$.........(i)

Step 2

Slope of $ AC=\large\frac{0-1}{1-0}$$=-1$

Since the altitude through $B$ is $\perp$ to $BC$,

Slope of the altitude through $B$ is $1$

Equation of the altitude through $B$ is given by

$y-0=1(x+1)$

$y-1=x$.......(ii)

Hence we have two altitude equations $x=0$ & $y-1=x$

Step 3

Orthocentre is the intersecting point of the altitudes (i) and (ii).

Solving (i) and (ii) we get $x=0$ and $y=1$

$\therefore$ The orthocentre is $(0,1)$

Hence (b) is the correct answer.

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