Intersection 81611

Given a triangle ABC: A (-1,3), B(2,-2), C(-4,-3). Determine the coordinates of the intersection of the heights and the coordinates of the intersection of the axes of the sides.

Correct answer:

x₁ = -0.4545
y₁ = 2.9091
x₂ = -1.2727
y₂ = -2.4545

Step-by-step explanation:

A = (- 1, 3) B = (2, - 2) C = (- 4, - 3) m_{1} = \frac{B _{y} - A _{y}}{B _{x} - A _{x}} = \frac{( - 2 ) - 3}{2 - ( - 1 )} = - \frac{5}{3} = - 1 \frac{2}{3} ≐ - 1.6667 m_{2} = \frac{B _{y} - C _{y}}{B _{x} - C _{x}} = \frac{( - 2 ) - ( - 3 )}{2 - ( - 4 )} = \frac{1}{6} ≐ 0.1667 m_{3} = \frac{A _{y} - C _{y}}{A _{x} - C _{x}} = \frac{3 - ( - 3 )}{( - 1 ) - ( - 4 )} = 2 A B ⊥ h_{1} M_{1} = - 1 / m_{1} = - 1 / (- 1.6667) = \frac{3}{5} = 0.6 M_{2} = - 1 / m_{2} = - 1 / 0.1667 = - 6 M_{3} = - 1 / m_{3} = - 1 / 2 = - \frac{1}{2} = - 0.5 h_{1} : (x_{1} - C_{x}) = M_{1} \cdot (y_{1} - C_{y}) (x_{1} - A_{x}) = M_{2} \cdot (y_{1} - A_{y}) (x_{1} - (- 4)) = 0.59999999999999 \cdot (y_{1} - (- 3)) (x_{1} - (- 1)) = (- 5.9999999999999) \cdot (y_{1} - 3) x_{1} - 0.6 y_{1} = - 2.2 x_{1} + 6 y_{1} = 17 R o w 2 - R o w 1 \to R o w 2 x_{1} - 0.6 y_{1} = - 2.2 6.6 y_{1} = 19.2 y_{1} = \frac{1 9 . 2}{6 . 6} = 2.90909091 x_{1} = - 2.2 + 0.59999999999999 y_{1} = - 2.2 + 0.6 \cdot 2.90909091 = - 0.45454545 = - 0.4545 x_{1} = \frac{- 5}{1 1} ≐ - 0.454545 y_{1} = \frac{3 2}{1 1} ≐ 2.909091 Verifying Solution: T_{3} = \frac{x _{1} - B _{x}}{y _{1} - B _{y}} = \frac{( - 0 . 4 5 4 5 ) - 2}{2 . 9 0 9 1 - ( - 2 )} = - \frac{1}{2} = - 0.5 T_{3} = M_{3}

y_{1} = = 2.9091

s_{1} = \frac{A _{x} + B _{x}}{2} = \frac{( - 1 ) + 2}{2} = \frac{1}{2} = 0.5 s_{2} = \frac{A _{y} + B _{y}}{2} = \frac{3 + ( - 2 )}{2} = \frac{1}{2} = 0.5 s_{3} = \frac{B _{x} + C _{x}}{2} = \frac{2 + ( - 4 )}{2} = - 1 s_{4} = \frac{B _{y} + C _{y}}{2} = \frac{( - 2 ) + ( - 3 )}{2} = - \frac{5}{2} = - 2 \frac{1}{2} = - 2.5 s_{5} = \frac{A _{x} + C _{x}}{2} = \frac{( - 1 ) + ( - 4 )}{2} = - \frac{5}{2} = - 2 \frac{1}{2} = - 2.5 s_{6} = \frac{A _{y} + C _{y}}{2} = \frac{3 + ( - 3 )}{2} = 0 (x_{2} - s_{1}) = M_{1} \cdot (y_{2} - s_{2}) (x_{2} - s_{3}) = M_{2} \cdot (y_{2} - s_{4}) (x_{2} - 0.5) = 0.59999999999999 \cdot (y_{2} - 0.5) (x_{2} - (- 1)) = (- 5.9999999999999) \cdot (y_{2} - (- 2.5)) x_{2} - 0.6 y_{2} = 0.2 x_{2} + 6 y_{2} = - 16 P i v o t : R o w 1 \leftrightarrow R o w 2 x_{2} + 6 y_{2} = - 16 x_{2} - 0.6 y_{2} = 0.2 R o w 2 - \frac{1}{1} \cdot R o w 1 \to R o w 2 x_{2} + 6 y_{2} = - 16 - 6.6 y_{2} = 16.2 y_{2} = \frac{1 6 . 2}{- 6 . 6} = - 2.45454545 x_{2} = \frac{- 1 6 - 5 . 9 9 9 9 9 9 9 9 9 9 9 9 9 y _{2}}{1} = \frac{- 1 6 - 6 \cdot ( - 2 . 4 5 4 5 4 5 4 5 )}{1} = - 1.27272727 = - 1.2727 x_{2} = \frac{- 1 4}{1 1} ≐ - 1.272727 y_{2} = \frac{- 2 7}{1 1} ≐ - 2.454545 Verifying Solution: S_{3} = \frac{x _{2} - s _{5}}{y _{2} - s _{6}} = \frac{( - 1 . 2 7 2 7 ) - ( - 2 . 5 )}{( - 2 . 4 5 4 5 ) - 0} = - \frac{1}{2} = - 0.5 S_{3} = M_{3}

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Dr. Math

To solve the problem, we need to find two specific points in triangle ABC with vertices at A(-1, 3) , B(2, -2) , and C(-4, -3) :

1. Orthocenter (Intersection of the Heights):

The orthocenter is the point where the three altitudes of the triangle intersect. An altitude is a perpendicular line from a vertex to the opposite side.

- Step 1: Find the slope of side BC :

Slope of BC = -3 - (-2)/-4 - 2 = -1/-6 = 1/6

The slope of the altitude from A to BC is the negative reciprocal:

Slope of altitude from A = -6

The equation of the altitude from A is:

y - 3 = -6(x + 1) ⇒ y = -6x - 3

- Step 2: Find the slope of side AC :

Slope of AC = -3 - 3/-4 - (-1) = -6/-3 = 2

The slope of the altitude from B to AC is the negative reciprocal:

Slope of altitude from B = -1/2

The equation of the altitude from B is:

y + 2 = -1/2(x - 2) ⇒ y = -1/2x - 1

- Step 3: Find the intersection of the two altitudes:

-6x - 3 = -1/2x - 1 ⇒ -12x - 6 = -x - 2 ⇒ -11x = 4 ⇒ x = -4/11

Substituting back:

y = -6(-4/11) - 3 = 24/11 - 33/11 = -9/11

So, the orthocenter is at:

(-4/11, -9/11)

2. Circumcenter (Intersection of the Perpendicular Bisectors of the Sides):

The circumcenter is the point where the perpendicular bisectors of the sides of the triangle intersect. It is the center of the circumcircle (the circle passing through all three vertices).

- Step 1: Find the midpoint of side AB :

Midpoint of AB = (-1 + 2/2, 3 + (-2)/2) = (1/2, 1/2)

The slope of AB is:

Slope of AB = -2 - 3/2 - (-1) = -5/3

The slope of the perpendicular bisector of AB is the negative reciprocal:

Slope of perpendicular bisector of AB = 3/5

The equation of the perpendicular bisector of AB is:

y - 1/2 = 3/5(x - 1/2) ⇒ y = 3/5x + 1/5

- Step 2: Find the midpoint of side BC :

Midpoint of BC = (2 + (-4)/2, -2 + (-3)/2) = (-1, -2.5)

The slope of BC is:

Slope of BC = 1/6

The slope of the perpendicular bisector of BC is the negative reciprocal:

Slope of perpendicular bisector of BC = -6

The equation of the perpendicular bisector of BC is:

y + 2.5 = -6(x + 1) ⇒ y = -6x - 8.5

- Step 3: Find the intersection of the two perpendicular bisectors:

3/5x + 1/5 = -6x - 8.5 ⇒ 3x + 1 = -30x - 42.5 ⇒ 33x = -43.5 ⇒ x = -43.5/33 = -29/22

Substituting back:

y = 3/5(-29/22) + 1/5 = -87/110 + 22/110 = -65/110 = -13/22

So, the circumcenter is at:

(-29/22, -13/22)

Final Answers:
- Orthocenter: (-4/11, -9/11)
- Circumcenter: (-29/22, -13/22)

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