Coordinates

Determine the coordinates of the vertices and the area 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 answer:

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

Step-by-step explanation:

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

Row 2 - 2/8 · Row 1 → Row 2
8x₀+3y₀ = -1
0.25y₀ = 1.25

y0 = 1.25/0.25 = 5
x0 = -1-3y0/8 = -1-3 · 5/8 = -2

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

Row 2 - 3/8 · Row 1 → Row 2
8x₁+3y₁ = -1
0.88y₁ = -2.63

y1 = -2.625/0.875 = -3
x1 = -1-3y1/8 = -1-3 · (-3)/8 = 1

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

Pivot: Row 1 ↔ Row 2
3x₂+2y₂ = -3
2x₂+y₂ = 1

Row 2 - 2/3 · Row 1 → Row 2
3x₂+2y₂ = -3
-0.33y₂ = 3

y2 = 3/-0.33333333 = -9
x2 = -3-2y2/3 = -3-2 · (-9)/3 = 5

x₂ = 5
y₂ = -9

Our linear equations calculator calculates it.

x = \frac{x _{2} + x _{1}}{2} = \frac{5 + 1}{2} = 3 y = \frac{y _{2} + y _{1}}{2} = \frac{( - 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

a = (x_{0} - x_{1})^{2} + (y_{0} - y_{1})^{2} = ((- 2) - 1)^{2} + (5 - (- 3))^{2} = 73 ≐ 8.544 h = ∣ p, C ∣ h = \frac{8 \cdot x _{2} + 3 \cdot y _{2} + 1}{8 ^{2} + 3 ^{2}} = \frac{8 \cdot 5 + 3 \cdot ( - 9 ) + 1}{8 ^{2} + 3 ^{2}} ≐ 1.6386 S = a \cdot h = 8.544 \cdot 1.6386 = 14

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