Coordinate System and Coordinates
14.0 Orthocentre of a triangle
14.0 Orthocentre of a triangle
The point of intersection of altitudes (i.e., the lines through the vertices and perpendicular to the opposite sides) is called the orthocenter.
As shown in figure, $ABC$ be the triangle with the vertices $A({x_1},{y_1})$, $B({x_2},{y_2})$ and $C({x_3},{y_3})$ whose sides $BC$, $CA$ and $AB$ are of lengths $a$, $b$ and $c$ respectively. $AD$, $BE$ and $CF$ are the perpendicular bisectors of sides $BC$, $CA$ and $AB$ respectively, then the co-ordinates of orthocentre of $\Delta ABC$ is $$\begin{equation} \begin{aligned} (\frac{{{x_1}\tan A + {x_2}\tan B + {x_3}\tan C}}{{\tan A + \tan B + \tan C}},\frac{{{y_1}\tan A + {y_2}\tan B + {y_3}\tan C}}{{\tan A + \tan B + \tan C}}) \\ or \\ (\frac{{a{x_1}\sec A + b{x_2}\sec B + c{x_3}\sec C}}{{a\sec A + b\sec B + c\sec C}},\frac{{a{y_1}\sec A + b{y_2}\sec B + c{y_3}\sec C}}{{a\sec A + b\sec B + c\sec C}}) \\\end{aligned} \end{equation} $$
The main difference between orthocentre and circumcentre is that in orthocentre the perpendicular from the vertex to the opposite side do not bisect the side means do not divide the side into two equal parts.
Note:
- Orthocenter of the obtuse angled triangle lies outside the triangle.
- Orthocenter of the right angled triangle is at the right angled vertex.
- Orthocenter, Centroid, and Circumcentre are always collinear and centroid divides the line joining orthocentre and circumcentre in the ratio $2:1$.
Question 4. If a triangle has its orthocentre at $(1,1)$ and circumcentre at $(\frac{3}{2},\frac{3}{4})$ then find the centroid of the triangle.
Solution: Since the centroid divides the orthocentre and circumcentre in the ratio $2:1$ (internally) and if centroid $G(x,y)$, then apply section formulae we get,
$$\begin{equation} \begin{aligned} x = \frac{{2 \times \frac{3}{2} + 1 \times 1}}{{2 + 1}} = \frac{4}{3} \\ y = \frac{{2 \times \frac{3}{4} + 1 \times 1}}{{2 + 1}} = \frac{5}{6} \\\end{aligned} \end{equation} $$
Therefore, centroid is $$(\frac{4}{3},\frac{5}{6})$$
Question 5. Show that the points $(a,0)$, $(0,b)$ and $(1,1)$ are collinear, if $\frac{1}{a} + \frac{1}{b} = 1$.
Solution: Let $A \equiv (a,0),{\text{ }}B \equiv (0,b){\text{ and }}C \equiv (1,1)$. The points are collinear if the area of triangle $ABC=0$ i.e.,
\[\begin{gathered}\frac{1}{2}\left| {\left| {\begin{array}{c}a&0 \\ 0&b \end{array}} \right| + \left| {\begin{array}{c}0&b \\ 1&1 \end{array}} \right| + \left| {\begin{array}{c}1&1 \\ a&0 \end{array}} \right|} \right| = 0 \hspace{1em} \\\Rightarrow \left| {(ab - 0) + (0 - b) + (0 - a)} \right| = 0 \hspace{1em} \\ab - a - b = 0 \hspace{1em} \\a + b = ab \hspace{1em} \\ \frac{1}{a} + \frac{1}{b} = 1 \hspace{1em} \\ \end{gathered} \]
