Question

Help will Give Medal


△ABC has vertices of A (0, 0), B (4, 4), and C (8, 0). Find the orthocenter of △ABC.

A.(0, 0)
B.;(4, 4)
C.(6, 2)
D.(3, 1)

Answer 1

Can one of you help me?

Answer 2

i cant see your drawing :/

Answer 3

can anyone else help?

Answer 4

I'll show examples, You can solve. OK. :)))

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

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-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

Answer 5

as an example.. :))