# Math Help - Find the orthocenter of the triangle

1. ## Find the orthocenter of the triangle

Points A(0,8), B(12, -8), C(-12,-8)

Slope of AB is $- \frac {4}{3}$

Slope of BC is is 0

Slope of AC is $\frac {4}{3}$

Perpindicular formula:

$m2= -\frac {1}{m1}$

m1 and m2 must not be = 0

Finding the perpindicular of $y= - \frac {4}{3} + b$ to point (12, -8)

Let m1 = $- \frac {4}{3}$

$m2 = - \frac {1}{-\frac {4}{3}} = \frac {3}{4}$

Solve for b

Let (x,y) = (12, -8)

$-8 = -\frac {3}{4}(12) + b$

$-8 = (9) + b$

Subtract both sides:

$1 = b$

Equation:

$y = -\frac {3}{4}x + 1$

Now for $y = \frac {3}{4}x + b$and (-12, -8)

is $y = \frac {3}{4}x + 1$

What else I need to do to get the Orthocenter of the triangle?

2. I don't know what you have done yet (your post is a bit chaotic put simply), I'll just say that you basically need the equation of two of the lines that cleanly cut one point of the triangle and make a right angle with the opposite side of the triangle. Once you get both equations of lines, solve the linear system to get the point of intersection of the two lines : that is the orthocenter of the triangle.

3. Originally Posted by Bacterius
I don't know what you have done yet (your post is a bit chaotic put simply), I'll just say that you basically need the equation of two of the lines that cleanly cut one point of the triangle and make a right angle with the opposite side of the triangle. Once you get both equations of lines, solve the linear system to get the point of intersection of the two lines : that is the orthocenter of the triangle.

i don't get it where to find those p and q

what i did was finding the slope of two lines...

4. By p and q I assume you mean the y-intercepts of each line. Each point of your triangle has a coordinate, right ? Then, you have the gradient of both lines, and you also have one point on each line. Therefore, if your line equations were :

$y = ax + p$
$y' = bx' + q$

You know (x, y) and (x', y'), and you also know the gradients a and b. Therefore :

$p = y - ax$
$q = y' - bx'$

5. Originally Posted by Bacterius
By p and q I assume you mean the y-intercepts of each line. Each point of your triangle has a coordinate, right ? Then, you have the gradient of both lines, and you also have one point on each line. Therefore, if your line equations were :

$y = ax + p$
$y' = bx' + q$

You know (x, y) and (x', y'), and you also know the gradients a and b. Therefore :

$p = y - ax$
$q = y' - bx'$

Ok here is what i got:
A(0,8), B(12,-8) C(-12,-8)
Line AB and AC are perpedicular

Midpoint of AB (6,0)
Slope-intercept of AB is -4/3
Perp-slope of AB is 3/4
perp-bisector of AB is

y-0=3/4(x-6)
y-0=3/4x - 4.5
y=3/4x -4.5

Midpoint of AC is (-6,0)
Slope-interceptor AC is 4/3
perp-slope of AC is -3/4
perp-bisector is

y-0=-3/4(x-(-6))
y-0=-3/4x+4.5
y=-3/4x+4.5

equate both AB and AC:
3/4x - 4.5 = -3/4x + 4.5

Get rid of fraction i multiply by 4
3x-18=3x+18

6x-18=18
6x=36
x=6

What else am i missing?

6. Are you looking for the orthocenter of ABC ? Sorry for all those questions but it is really important to explain your problem in detail. No one likes to be told that "nah I wasn't talking about that".

7. Originally Posted by Bacterius
Are you looking for the orthocenter of ABC ? Sorry for all those questions but it is really important to explain your problem in detail. No one likes to be told that "nah I wasn't talking about that".
Yes im trying to get the orthocenter and I don't if im going there.

Line AB equation is y= 4/3x + 8

Line AC equation is y= -4/3 + 8

I really don't know how to get there.

8. There is something wrong with your question. If AB and AC are perpendicular, then ABC is a right-angled triangle, in which case the coordinates of its orthocenter are trivial (equal to the point with the right angle, here A).

9. Originally Posted by Anemori
Points A(0,8), B(12, -8), C(-12,-8)

Slope of AB is $- \frac {4}{3}$

Slope of BC is is 0

Slope of AC is $\frac {4}{3}$

Perpindicular formula:

$m2= -\frac {1}{m1}$

m1 and m2 must not be = 0

Finding the perpindicular of $y= - \frac {4}{3}\color{red}x \color{black}+ b$ to point (-12, -8)

Let m1 = $- \frac {4}{3}$

$m2 = - \frac {1}{-\frac {4}{3}} = \frac {3}{4}$

Solve for b

Let (x,y) = (12, -8)

$-8 = -\frac {3}{4}(12) + b$
$-8 = \color{red}-\color{black}(9) + b$

Subtract both sides:

$1 = b$

Equation:

$y = -\frac {3}{4}x + 1$

Now for $y = \frac {3}{4}x + b$and (-12, -8)

is $y = \frac {3}{4}x + 1$

What else I need to do to get the Orthocenter of the triangle?
You must now find the point of intersection of the 2 lines
that you have written,
the perpendiculars to 2 sides that shoot through the opposite vertex.

You need the point of intersection of

$y=-\frac{3}{4}x+1$

and

$y=\frac{3}{4}x+1$

If x=0, y=1 in both cases, so i guess that's it!

It will be very clear if you draw a little sketch.

10. Originally Posted by Anemori
Yes im trying to get the orthocenter and I don't if im going there.

Line AB equation is y= 4/3x + 8

Line AC equation is y= -4/3 + 8

I really don't know how to get there.
ok this is what i got again:

Slope-intercept is 4/3 and perp-slope is -3/4 of AB
Slope-intercept is -4/3 and perp-slope is 3/4 of AC

then this what i did.

Line AB, Let (x,y) = B(12,-8)

y-y1 = m(x-x1)
y-(-8)=-3/4(x-12)
y+8=-3/4x+9
y=-3/4x+1

Line AC, Let (x,y) = C(-12,-8)
y-y1=m(x-x1)
y-(-8)=3/4(x-(-12))
y+8=3/4x(x+12)
y+8=3/4x+9
y=3/4x+1

I Equate both to get the point coor of orthocenter:

-3/4x+1=3/4x+1
multiply both fraction by 4:
-3x+1=3x+1
-6x+1=1
-6x=0
x=0

Point coor of orthocenter is (0,1) am i right?

11. Originally Posted by Anemori
ok this is what i got again:

Slope-intercept of AB is -4/3 and perp-slope is -3/4 of AB
Slope-intercept of AC is -4/3 and perp-slope is -3/4 of AC

then this what i did.

Line AB, Let (x,y) = B C(-12,-8) C lies on the perpendicular to AB

y-y1 = m(x-x1)
y-(-8)=- 3/4(x--12)
y+8=- 3/4x+9
y=- 3/4x+1

Line AC, Let (x,y) = C B(-12,-8) B lies on the perpendicular to AC
y-y1=m(x-x1)
y-(-8)=-3/4(x-(-12))
y+8=-3/4x(x+-12)
y+8=-3/4x+9
y=-3/4x+1

I Equate both to get the point coor of orthocenter:

-3/4x+1=3/4x+1
multiply both fraction lines by 4:
-3x+1 4=3x+1 4
-6x+1 4=1 4
-6x=0
x=0

Point coor of orthocenter is (0,1) am i right? Yes, coincidentally
Hi Anemori,

You have the idea,
but you have some work to do yet.

You've got to co-ordinate your text better.
You got the right answer due to the fact that your errors cancelled each other out.

If you draw the points, lines and perpendiculars, you will see why.

Please study the attachment visually, as you are working in the dark.

Choosing the y-axis as one of the perpendicular lines is a shortcut.