Find the co-ordinates of the orthocentre of triangle ABC whose vertices are A(1,2), B(2,3) and C(4,3)

Slope of AB=$\frac{3-2}{2-1}=1$

Slope of BC=$\frac{3-3}{4-2}=0$

Slope of CA=$\frac{3-2}{4-1}=\frac{1}{3}$

Let AD, BE and CF be the altitudes in $△ABC$

Then,

Slope of AD=$-\frac{1}{Slope of BC}=-\frac{1}{0}$

Slope of BE=$-\frac{1}{Slope of CA}=-\frac{1}{{\displaystyle \frac{1}{3}}}=-3$

Slope of CF=$-\frac{1}{Slope of AB}=-\frac{1}{1}=-1$

We now have the vertices and slopes of AD, BE and CF. Let (x, y) be the orthocentre. So, by the equation $y-y_1=m\left(x-x_1\right)$, we have:-

$$\begin{gathered} AD: y-2=-\frac{1}{0}\left(x-1\right) \Rightarrow x=1 [the line is vertical] \\ BE: y-3=-3\left(x-2\right) \Rightarrow y-3=-3x+6 \\ CF: y-3=-1\left(x-4\right) \Rightarrow y-3=-x+4 \end{gathered}$$

Solving these equations, we find x=1, y=6

Thus, the required orthocentre is (1, 6).