Orthocenter: Definition and How to Find with Example

In this article, we will cover the concept of Orthocentre. This category falls under the broader category of Coordinate Geometry, which is a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination(JEE Main) and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. A total of thirteen questions have been asked on JEE MAINS( 2013 to 2023) from this topic including one in 2018, one in 2019, two in 2021, five in 2022, and four in 2023.

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Orthocentre: Definition

The Orthocentre (H) of a triangle is the point of intersection of altitudes that are drawn from one vertex to the opposite side of a triangle.

The point of concurrency of altitudes of a triangle is called the orthocentre of the triangle. The Orthocentre of the acute-angled triangle lies inside of the triangle and that of the obtuse-angled triangle lies outside the triangle. The Orthocentre of a right-angled triangle is the vertex where the right angle occurs.

Orthocentre of Triangle Formula

Let $\mathrm{A}\left(\mathrm{x}_1, \mathrm{y}_1\right), \mathrm{B}\left(\mathrm{x}_2, \mathrm{y}_2\right)$, and $\mathrm{C}\left(\mathrm{x}_3, \mathrm{y}_3\right)$ be the vertices of the triangle ABC, then coordinates of orthocentre are:

Coordinates of Orthocentre (H) is

$\left(\frac{\mathbf{x}_1 \tan \mathbf{A}+\mathbf{x}_{\mathbf{2}} \tan \mathbf{B}+\mathbf{x}_{\mathbf{3}} \tan \mathbf{C}}{\tan \mathbf{A}+\tan \mathbf{B}+\tan \mathbf{C}}, \frac{\mathbf{y}_{\mathbf{1}} \tan \mathbf{A}+\mathbf{y}_{\mathbf{2}} \tan \mathbf{B}+\mathbf{y}_{\mathbf{3}} \tan \mathbf{C}}{\tan \mathbf{A}+\tan \mathbf{B}+\tan \mathbf{C}}\right)$

Properties of Orthocentre

1) For a right-angled triangle, the orthocenter is the vertex containing the right-angle

2) In a triangle other than the equilateral triangle, the orthocentre (H), centroid (G), and circumcentre (0) are collinear with a ratio HG: G O = 2: 1

3) If the vertices of the triangle are $\left(x_1, y_1\right)\left(x_2, y_2\right)$ and $\left(x_3, y_3\right)$ and the circumcentre is (0, 0), then the orthocentre is $\left(x_1+x_2+x_3, y_1+y_2+y_3\right)$).

Solved Examples Based on Orthocentre

Example 1: If $(\alpha, \beta)$ is the orthocenter of the triangle ABC with vertices A (3, –7), B(–1,2), and C (4, 5), then 9$(\alpha)$ – 6$(\beta)$ + 60 is equal to [JEE MAINS 2023]

Solution

Altitude of $B C=y+7=\frac{-5}{3}(x-3)$
$
\begin{aligned}
& 3 y+21=-5 x+15 \\
& 5 x+3 y+6=0
\end{aligned}
$

Altitude of $\mathrm{AC}: y-2=\frac{-1}{12}(\mathrm{x}+1)$
$
\begin{aligned}
& 12 y-24=-x-1 \\
& x+12 y=23 \\
& \alpha=\frac{-47}{19}, \quad \beta=\frac{121}{57} \\
& 9 a-6 \beta+60=25
\end{aligned}
$

Hence, the answer is 25

Example 2: If the orthocentre of the triangle, whose vertices are $(1,2)(2,3)$ and $(3,1)$ is $(\alpha, \beta)$, then the quadratic equation whose roots are $\alpha+4 \beta$ and $4 \alpha+\beta$, is [JEE MAINS 2023]

Solution

equation of $A D: x-2 y+3=0$

equation of $C E: x+y-4=0$

orthocenter $(\alpha, \beta)$ is $\left(\frac{5}{3}, \frac{7}{3}\right) \alpha+4 \beta=11$ and $4 \alpha+\beta=9$

Quadratic equation is$\begin{aligned} & x^2-(11+9) x+(11 \times 9)=0 \\ & \Rightarrow x^2-20 x+99=0\end{aligned}$

Hence, the answer is $x^2-20 x+99=0$

Example 3: For $\mathrm{t} \in(0,2 \pi)$ if ABC is an equilateral triangle with vertices $A(\sin t,-\cos t), B(\cos t, \sin t)$ and $C(a, b)$ such that its orthocentre lies on a circle with a center $\left(1, \frac{1}{3}\right)$then $\left(a^2-b^2\right)$ is equal to: [JEE MAINS 2022]

Solution: $\mathrm{s} \equiv \sin \mathrm{t}, \mathrm{c} \equiv \cos \mathrm{t}$
Let orthocentre be $(h, k)$
since it if an equilateral triangle hence orthocentre cor coincides with centroid$\therefore \mathrm{a}+\mathrm{s}+\mathrm{c}=3 \mathrm{~h}, \mathrm{~b}+\mathrm{s}-\mathrm{c}=3 \mathrm{k}$

$\begin{aligned} & \quad \therefore(3 h-a)^2+(3 k-b)^2=(s+c)^2+(s-c)^2=2\left(s^2+c^2\right)=2 \\ & \therefore\left(k-\frac{a}{3}\right)^2+\left(k-\frac{b}{3}\right)^2=\frac{2}{4}\end{aligned}$

$\begin{aligned} & \text { circle center at }\left(\frac{a}{3}, \frac{b}{3}\right) \\ & \text { gives, } \frac{a}{3}=1, \frac{b}{3}=\frac{1}{3} \Rightarrow a=3, b=1 \\ & \therefore a^2+a^2-b^2=8\end{aligned}$

Hence, the answer is 8

Summary

The orthocenter, where altitudes intersect, reflects the triangle’s perpendicular relationships, while the circumcenter, at the intersection of perpendicular bisectors, defines the centre of the circumcircle. Knowledge of these concepts enriches understanding of triangle symmetry, circle constructions, and their practical applications in fields ranging from architecture to mathematical problem-solving.