# Find the coordinates of the orthocenter of the triangle formed by straight lines?

## $x - y - 5 = 0$ , $2 x - y - 8 = 0$ and $3 x - y - 9 = 0$

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#### Explanation

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#### Explanation:

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May 2, 2017

The orthocentre is at (–6, 1).

#### Explanation:

To find the orthocentre of a triangle, we connect each vertex to its opposite side with a perpendicular line. All three of these lines will intersect at a single point, called the orthocentre.

(Note that for obtuse triangles, the orthocentre will be outside the triangle.)

Let the triangle's vertices be $A$, $B$, and $C$. Our process will involve 3 steps:

1. Find the equation of a line that passes through $C$, and is perpendicular to $\overline{A B}$.
2. Find the equation of a line that passes through $B$, and is perpendicular to $\overline{A C}$.
3. Solve the system of linear equations made from the equations in steps 1 and 2.

(We don't need to bother with finding the third line, since it will be redundant.)

Uh-oh—looks like we don't know the coordinates of the vertices. We'll have to find two of them first.

Let's call the line given by the first equation "side $a$", and the middle equation "side $b$". Solving this system of two equations gives

$\textcolor{w h i t e}{\implies} y = x - 5 = 2 x - 8$
$\implies \textcolor{w h i t e}{y = x -} 3 = x$

$\implies y = 3 - 5$
color(white)(=>y)=–2

Thus, $a$ and $b$ intersect at C=(3, –2).

The last equation gives us side $c$ (also called $\overline{A B}$). We now find the line that passes through (3, –2) and is perpendicular to $3 x - y - 9 = 0$.

Perpendicular lines have slopes that are negative reciprocals of each other. (Here's an explanation why: https://socratic.org/featured?s=365506) The slope of $3 x - y - 9 = 0$ is 3 (all we do is move the $y$ to the right, and the equation is in $y = m x + b$ form). Thus, the slope of a perpendicular line to this one is $- \frac{1}{3}$.

We now want to find the equation of a line passing through (x, y)=(3, –2) with a slope of m=–1/3. Plug these into the equation of a line to find the necessary $y$-intercept:

$\text{ "y=" "mx" } + b$
–2=–1/3(3)+b
–2=" "-1"   "+b
–1=b

So the equation of the altitude from $\overline{A B}$ to $C$ is y=–1/3x-1. Step 1 is done!

For brevity, I'll leave step 2 to you as an exercise. The equation is y=–1/2x-2.

Now, all we do is find where these two lines meet (step 3). Set the equations equal to each other and solve:

y=–1/3x-1=–1/2x-2

color(white)(y=)"  "–2x-6=–3x-12 (multiply through by 6)

"                      "x=–6

=>y=–1/3(–6)-1

$\textcolor{w h i t e}{\implies} y = 2 - 1$
$\textcolor{w h i t e}{\implies} y = 1$

There we go! After all that, the orthocentre of the triangle is at (x, y)=(–6, 1).

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