Triangle ABC has vertices A(0, 6) B(4, 6) C(1, 3). Sketch a graph of ABC and use it to find the orthocenter of ABC.

Solution:

Given triangle with vertices ABC , A(0, 6) B(4, 6) C(1, 3)

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)

We find slopes of AB, BC,CA {Slope formula y - y’⁄ x - x’}}

AB = 6 - 6/4 - 0 = 0/4 = 0

BC = 3 - 6/ 1 - 4 = -3/-3 = 1

CA = 6 - 3/ 0 - 1 = 3/-1 =-3

But we know Orthocentre is the point where perpendiculars 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

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

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

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 = m(x - X)

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) ortho center

(x, y) = (1, 5) orthocentre.

---

Triangle ABC has vertices A(0, 6) B(4, 6) C(1, 3). Sketch a graph of ABC and use it to find the orthocenter of ABC.

Summary:

Triangle ABC has vertices A(0, 6) B(4, 6) C(1, 3), a graph of ABC is sketched and the orthocenter is (1, 5).