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

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

##### 1 Answer

#### Answer

#### Answer:

#### Explanation

#### Explanation:

#### Answer:

The orthocentre is at

#### 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

- Find the equation of a line that passes through
#C# , and is perpendicular to#bar(AB)# . - Find the equation of a line that passes through
#B# , and is perpendicular to#bar(AC)# . - 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

#color(white)(=>)y=x-5=2x-8#

#=>color(white)(y=x-)3=x#

#=>y=3-5#

#color(white)(=>y)=–2#

Thus,

The last equation gives us side

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 *perpendicular* line to this one is

We now want to find the equation of a line passing through

#" "y=" "mx" "+b#

#–2=–1/3(3)+b#

#–2=" "-1" "+b#

#–1=b#

So the equation of the altitude from

For brevity, I'll leave step 2 to you as an exercise. The equation is

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#

#color(white)(=>)y=2-1#

#color(white)(=>)y=1#

There we go! After all that, the orthocentre of the triangle is at

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