ABC has vertices A(0, 6), B(4, 6), and C(1, 3). Find the orthocenter of ABC. List your steps.

Question

please please please help me!!!

Answer 1

The point where the altitudes of a triangle meet called Ortho Centre

We have given a triangle ABC whose vertices are(0, 6),(4, 6), (1, 3)

In Step 1                we find slopes  Of AB, BC,CA           Slope formulae y_2-y_1⁄  x2-X1

slope    AB= 6-6/4-0  = 0/4   =0

 ....      BC= 3-6/ 1-4 = -3/-3 =1

.......    CA=6-3/ 0-1 =3/-1   =-3

In Step 2

But we know Orthocentre is the point where perpendeculars drawn from vertex to opposite side meet.  So

Let's think a triangle ABC and AD, BE, CF are perpendiculars drawn to the vertex.

Slope AD = -1/slope BC    = -1/1 =-1

.......BE = -1/slope CA      = -1/-3 = 1/3

.....CF  = -1/slope AB      = -1/0   undefined

Step 3   we have now vertices and slopes of AD,BE,CF  we find equations of lines AD,BE and CF

  we have A(0,6) and m =-1 we substitute in the equation y_-y_{1 =} m(x-X₁₎

y-6=-1(x-0)

y+x=6     -  eq   1

B(4,6) and slope BE   (1/3)

y-6=1/3(x-4)

3Y-18=x-4

3y-x=14    -eq  2

C(1,3)   and whose slope CF undefined

 So line is vertical and x=1 is the eq

 

Now solving any of equations 1&2 we get values for( x,y) orhto centre      

(x,y)  =(

 solving eq 1 and eq 2 we get x=1, y=5    (

      I hope this helps you

 

 

 

Answer 2

Orthocenter is the intersection point of the three lines through a vertex perpendicular to the opposite side. For example, through A, perpendicular to BC; through B, perpendicular to AC; and through C, perpendicular AB.

Line through C perpendicular to AB:

m = (6-6) / (4-0) = 0

The line is horizontal, so a line perpendicular to that one would be vertical and have an undefined slope.

However, since it's going through C, whose x coordinate is 1, the equation for the line through C perpendicular to AB is:

x=1

Line perpendicular to BC through A:

m of BC = (6-3)/(4-1) = 3/3 = 1

m perpendicular to BC is therefore -1.

Line perp. to BC through A ... 6 = -1(0) + b

b = 6

Equation of line = y = -1(x) + 6.

Line perpendicular to AC through B:

m of AC = (6-3)/(0-1) = 3/-1 = -3.

m perp to AC = 1/3

Substituting in y=mx+b gives 6 = (1/3)(4) + b

b = 14/3

Line equation -> y = (1/3)x + (14/3)

Since x must be 1 for all three altitudes,

plug in 1 for the other two equations.

y = -(x) + 6

y = -(1) + 6 = 5

(1,5) works for the first two equations.

Verify in the 3rd.

y = (1/3)(1) + (14/3)

y = 15/3 = 5.

The orthocenter of this triangle is at (1,5).

Answer 3

Here is another way to do it: vector approach.

Since AB is a horizonal line, the altitude through C must be a vertical line. So, the orthocenter coordinates can be written as (1, b), and the direction of the altitude from A is (1, b-6), and the direction of BC is parallel to (1, 1). Since the dot product of two perpendicular vector is zero, we have

(1, b-6) ⋅ (1, 1) = 0

1 + b-6 = 0

b = 5

Answer: The orthocenter is at (1, 5).