Question

Triangle ABC has vertices A(0,0) B(3,3) C(6,0) find the orthcenter of triangle abc

Answer 1

Answer:

(3, 0)

Step-by-step explanation:

The orthocentre of a triangle is the intersection point of the three perpendicular bisectors of the three sides.

A perpendicular bisector cuts a line in half and meets it at 90°.

You need to know the midpoint and slope for at least two of the sides.

Midpoint formula: $M = (\frac{x_{2}+x_{1}}{2} ,\frac{y_{2}+y_{1}}{2})$

Slope formula: $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$

When a line a perpendicular to another, its slope is the negative reciprocal.

The midpoint of the side is a point on the bisector.

Let's focus on the side AB.

Info set 1: A(0,0) x₁ = 0  y₁ = 0

Info set 2: B(3,3) x₂ = 3 y₂ = 3

Find the midpoint:

$M = (\frac{x_{2}+x_{1}}{2} ,\frac{y_{2}+y_{1}}{2})$

$M = (\frac{3+0}{2} ,\frac{3+0}{2})$

$M = (\frac{3}{2} ,\frac{3}{2})$

$M = (1.5, 1.5)$

Find the slope:

$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$

$m = \frac{3 - 0}{3 - 0}$

$m = \frac{3}{3}$

$m = 1/1$

To find the perpendicular line's slope, find the negative reciprocal. (Switch the top and bottom of the slope in fraction form, then change the negatuve/positive)

⊥m = -1/1

⊥m = -1

Find the equation of ⊥AB using m = -1 and (1.5, 1.5).

Substitute into the equation of a line.

y = mx + b

1.5 = -1(1.5) + b

1.5 = -1.5 + b   Adding 1.5 to both sides

b = 3

⊥AB :  y = -x + 3

Do the same for side AC

Info set 1: A(0,0)  x₁ = 0  y₁ = 0

Info set 2: C(6,0)  x₂ = 6 y₂ = 0

Find the midpoint:

$M = (\frac{x_{2}+x_{1}}{2} ,\frac{y_{2}+y_{1}}{2})$

$M = (\frac{6+0}{2} ,\frac{0+0}{2})$

$M = (\frac{6}{2} ,\frac{0}{2})$

$M = (3, 0)$

Find the slope:

$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$

$m = \frac{0 - 0}{6 - 0}$

$m = \frac{0}{6}$

m = 0      This means the line is horizontal

A line perpendicular would be vertical. (⊥m = ∞)

Vertical lines are written in the form y=x, where x is the x-intercept.

Since the midpoint is (3, 0), the midpoint is 3.

⊥AC : y = 3

The intersection point of y = 3 and y = -x + 3 is the orthcentre.

Substitute y for 3 in the equation y = -x + 3

y = -x + 3

3 = -x + 3

x = 0

The orthocentre is (3, 0).

Check for the other line ⊥BC.

Info set 1: B(3,3) x₁ = 3  y₁ = 3

Info set 2: C(6,0)  x₂ = 6 y₂ = 0

Find the midpoint:

$M = (\frac{x_{2}+x_{1}}{2} ,\frac{y_{2}+y_{1}}{2})$

$M = (\frac{6+3}{2} ,\frac{0+3}{2})$

$M = (\frac{9}{2} ,\frac{3}{2})$

$M = (4.5 ,1.5)$

Find the slope:

$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$

$m = \frac{0- 3}{6 - 3}$

$m = \frac{-3}{3}$

$m = \frac{-1}{1}$

Find its negative reciprocal

⊥m = 1

Substitute slope and the point into the equation of a line

y = mx + b

1.5 = 1(4.5) + b

1.5 = 4.5 + b

b = -3

⊥BC : y = x - 3

Substitute the orthocentre (3,0)

y = x - 3

0 = 3 - 3

0 = 0

LS = RS

Therefore the orthocentre is (3, 0)