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ABC has vertices A(0, 6), B(4, 6), and C(1, 3). Find the orthocenter of ABC. List your steps.
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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 y2-y1⁄ 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
we have A(0,6) and m =-1 we substitute in the equation y-y1 = 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
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).
