Coordinates

Determine the coordinates of the vertices and the content of the parallelogram, the two sides of which lie on the lines 8x + 3y + 1 = 0, 2x + y-1 = 0 and the diagonal on the line 3x + 2y + 3 = 0

Correct result:

x₀ =  -2
y₀ =  5
x₁ =  1
y₁ =  -3
x₂ =  5
y₂ =  -9
x₃ =  8
y₃ =  -17
S =  14

Solution:

8x₀+3y₀+1=0
2x₀+y₀-1=0

8•x₀+3•y₀+1=0
2•x₀+y₀-1=0

8x₀+3y₀ = -1
2x₀+y₀ = 1

x₀ = -2
y₀ = 5

Our linear equations calculator calculates it.

8x₁+3y₁+1=0
3x₁+ 2y₁+3=0

8•x₁+3•y₁+1=0
3•x₁+ 2•y₁+3=0

8x₁+3y₁ = -1
3x₁+2y₁ = -3

x₁ = 1
y₁ = -3

Our linear equations calculator calculates it.

2x₂+y₂-1=0
3x₂+ 2y₂+3=0

2•x₂+y₂-1=0
3•x₂+ 2•y₂+3=0

2x₂+y₂ = 1
3x₂+2y₂ = -3

x₂ = 5
y₂ = -9

Our linear equations calculator calculates it.

$x=\dfrac{ x_{2}+x_{1} }{ 2 }=\dfrac{ 5+1 }{ 2 }=3 \ \\ y=\dfrac{ y_{2}+y_{1} }{ 2 }=\dfrac{ (-9)+(-3) }{ 2 }=-6 \ \\ \ \\ x_{3}=-x_{0}+2 \cdot \ x=-(-2)+2 \cdot \ 3=8$

$y_{3}=-y_{0}+2 \cdot \ y=-5+2 \cdot \ (-6)=-17$

$$\begin{gathered} a=\sqrt{(x_0-x_1{)}^2+(y_0-y_1{)}^2}=\sqrt{((-2)-1{)}^2+(5-(-3){)}^2}=\sqrt{73}\doteq 8.544 \\ h=\mathrm{\mid }p,C\mathrm{\mid } \\ h={\displaystyle \frac{8\cdot x_2+3\cdot y_2+1}{\sqrt{8^2+3^2}}}={\displaystyle \frac{8\cdot 5+3\cdot (-9)+1}{\sqrt{8^2+3^2}}}\doteq 1.6386 \\ S=a\cdot h=8.544\cdot 1.6386=14 \end{gathered}$$

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