In the following triangle \(▵ABC\),

\(AQ\), \(BR\) and \(CP\) are the **altitudes** of the triangle.

They intersect at the point \(H\).

This point \(H\) is the **orthocenter** of \(△ABC\).

**Note: **An altitude is a line segment that is drawn from
a vertex of a triangle perpendicular to the opposite side of that vertex.

A triangle has three altitudes because it has three sides.

- ✓ Orthocenter is the point of intersection of the altitudes.
- ✓ For acute triangles, it lies
**inside the triangle.** - ✓ For obtuse triangles, it lies
**outside the triangle.** - ✓ For right-angled triangles, it lies
**on the 90° vertex.** - ✓ In an equilateral triangle, the centroid, the
**orthocenter**, - the circumcenter and the incenter coincide on the same point.

Before finding the **coordinates of an orthocenter**,
let us check what is **slope of a perpendicular**.

Slope of a perpendicular line is the **negative reciprocal** of
the slope of the line to which it is perpendicular.

In the above, \(MN\) is perpendicular to \(OP\).
If slope of \(OP\) is \(\Large m\),
then slope of \(MN\) will be \(\Large-\frac{1}{m} \).

For an acute triangle,
the orthocenter lies
**inside the triangle**.

Note that we had to extend sides \(AC\) and \(BC\).

The perpendiculars \(AQ\), \(BR\) and \(PC\) are also
extended to meet at \(H\), which is the orthocenter of the triangle.

For an obtuse triangle,
the orthocenter always **lies outside the triangle**.

Side \(AB\) and \(⊥AQ\) coincided on each other.

Side \(BC\) and \(⊥CP\) coincided on each other.

Point \(H\) is the orthocenter of the triangle.

For a right-angled triangle, the orthocenter always
**lies on the \(90°\) vertex**.

If the coordinates of the vertices \(A\), \(B\) and \(C\) of a triangle are \( (x_1, y_1) \), \( (x_2, y_2) \) and \( (x_3, y_3) \) respectively, then the coordinates of the orthocenter \(H(x, y)\) can be calculated with the following steps:

- ✓ Find slopes of any two sides, say \(AB\) and \(BC\).
- ✓ Find slopes of altitudes(\(AQ\) and \(CP\)) on those sides .
- ✓ Find the two equations of the altitudes \(AQ\) and \(CP\).
- ✓ Solve \(x\) and \(y\) from the two equations.
- ✓ \(H(x, y)\) gives us the intersection point of \(AQ\) and \(CP\)

Slope of AB \(\; \large m_{AB} = \LARGE {\frac{y_2 - y_1}{x_2 - x_1} } \)

Slope of BC \(\; \large m_{BC} = \LARGE {\frac{y_3 - y_2}{x_3 - x_2} } \)

Slope of CP \(\; \large m_{CP} = \frac{-1}{m_{AB}} \quad \small\color{gray}[\;\because CP\perp AB \;] \)

Slope of AQ \(\; \large m_{AQ} = \frac{-1}{m_{BC}} \quad \small\color{gray}[\;\because AQ\perp BC \;] \)

Equation of AQ\(\quad \large y - y_1 = m_{AQ} (x - x_1) \quad\color{red}... eqn. (a)\)

Equation of CP\(\quad \large y - y_3 = m_{CP} (x - x_3) \quad\color{red}... eqn. (b)\)

Solving the above two equations gives us the values of \(x\) and \(y\).

Let us go with a \(△ABC\) whose coordinates are \(A(-30, 2)\), \(B(10, 4)\) and \(C(-4, 37)\).
We have to find the coordinates of the orthocenter point \(H(x,y)\).

Slope of \(AB :\)

\(\; \large m_{AB} = {\frac{4 - 2}{10 - (-30)}} = 0.05 \)

Slope of \(BC :\)

\(\; \large m_{BC} = {\frac{37 - 4}{-4 - 10} = -2.357} \)

Slope of \(CP :\)

\(\; \large m_{CP} = \frac{-1}{0.05} = -20\quad \color{gray} [\;\small\because CP\perp AB \;] \)

Slope of \(AQ :\)

\(\; \large m_{AQ} = \frac{-1}{-2.357} = 0.424\quad \color{gray} [\;\small\because AQ\perp BC \;] \)

Equation of perpendicular \(AQ :\)

\begin{align}
& y - 2 = 0.424\times(x - (-30)) \\\\
\Rightarrow\; & y - 2 = 0.424\times(x + 30) \\\\
\Rightarrow\; & y - 2 = 0.424\times x + 0.424\times 30 \\\\
\Rightarrow\; & y - 2 = 0.424x + 12.72 \\\\
\Rightarrow\; & 0.424x - y = -2-12.72 \\\\
\therefore\; & 0.424x - y = -14.72 \quad\color{red}... (i) \\\\
\end{align}
Equation of perpendicular \(CP :\)

\begin{align}
& y - 37 = -20 (x - (-4)) \\\\
\Rightarrow\; & y - 37 = -20 (x + 4) \\\\
\Rightarrow\; & y - 37 = -20 \times x - 20 \times 4 \\\\
\Rightarrow\; & y - 37 = -20x - 80 \\\\
\Rightarrow\; & 20x + y = 37 - 80 \\\\
\therefore\; & 20x + y = -43 \quad\color{red}... (ii) \\\\
\end{align}
By adding equation \(\color{red}(i)\) and \(\color{red}(ii) :\)

\begin{align}
& 20.424x = -57.72 \\\\\
& \therefore x = -2.826 \\\\
\end{align}
Putting the value of \(x\) in equation \(\color{red}(ii)\)

\begin{align}
& 20 \times (-2.826) + y = -43 \\\\\
& -56.52 + y = -43 \\\\\
& y = 56.52-43 \\\\\
& \therefore y = 13.52 \\\\\
& \therefore H(x, y) = (-2.826, 13.52) \\\\\
\end{align}
So, orthocenter of the above triangle is at \((-2.826, 13.52)\).

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