please please please help me!!!

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

# 3 Answers

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).

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-X_{1)}*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*

* *

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).