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




8x0+3y0 = -1
2x0+y0 = 1

x0 = -2
y0 = 5

Calculated by our linear equations calculator.

3x1+ 2y1+3=0

3•x1+ 2•y1+3=0

8x1+3y1 = -1
3x1+2y1 = -3

x1 = 1
y1 = -3

Calculated by our linear equations calculator.

3x2+ 2y2+3=0

3•x2+ 2•y2+3=0

2x2+y2 = 1
3x2+2y2 = -3

x2 = 5
y2 = -9

Calculated by our linear equations calculator.
x=x2+x12=5+12=3 y=y2+y12=(9)+(3)2=6  x3=x0+2 x=(2)+2 3=8x=\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
y3=y0+2 y=5+2 (6)=17y_{3}=-y_{0}+2 \cdot \ y=-5+2 \cdot \ (-6)=-17
a=(x0x1)2+(y0y1)2=((2)1)2+(5(3))2738.544 h=p,C h=8 x2+3 y2+182+32=8 5+3 (9)+182+321.6386 S=a h=8.544 1.6386=14a=\sqrt{ (x_{0}-x_{1})^2+(y_{0}-y_{1})^2 }=\sqrt{ ((-2)-1)^2+(5-(-3))^2 } \doteq \sqrt{ 73 } \doteq 8.544 \ \\ h=|p,C| \ \\ h=\dfrac{ 8 \cdot \ x_{2}+3 \cdot \ y_{2}+1 }{ \sqrt{ 8^2+3^2 } }=\dfrac{ 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

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