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}

Did you find an error or inaccuracy? Feel free to write us. Thank you!

Tips for related online calculators

The line slope calculator is helpful for basic calculations in analytic geometry. The coordinates of two points in the plane calculate slope, normal and parametric line equation(s), slope, directional angle, direction vector, the length of the segment, intersections of the coordinate axes, etc.
Our vector sum calculator can add two vectors given by their magnitudes and by included angle.
Do you have a linear equation or system of equations and are looking for its solution? Or do you have a quadratic equation?
See also our trigonometric triangle calculator.