If a(-2,1) ,b(9,0),c(4,6)and d(1, )are the vertex of parallelogram ABCD find the value of A and B hence find the length of the sides
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A parallelogram has two sets of of parallel lines. The length and the width.
We will use the relationship between parallel lines to get the point D.
Parallel lines have the same gradient.
AB // DC... That is they are parallel hence have the same gradient.
The formulae for getting the length :
L = √{(y₁ - y₂) ² + (x₁ - x₂) ²}
The formulae for getting the gradient :
(y₁ - y₂) /(x₁ - x₂)
CALCULATIONS :
Gradient for AB :
(0 - 1) / (9 - [-2]) = - 1 / 11
Gradient for DC = gradient for AB
Let point D = ( 1, y) then:
(6 - Y) / (4 - 1) = - 1/11
6 - y / 3 = - 1/11
11(6 - Y) = - 1 × 3
66 - 11y = - 3
- 11y = - 99
y = 9
LENGTH :
AB = DC
√(-1)² + (11)² = √122 = 11.0454 units
AB = DC = 11.0454units
DA = BC
√(6²) + (-5)² = √61 = 7.8102 units
DA = BC = 7.8102Units.
We will use the relationship between parallel lines to get the point D.
Parallel lines have the same gradient.
AB // DC... That is they are parallel hence have the same gradient.
The formulae for getting the length :
L = √{(y₁ - y₂) ² + (x₁ - x₂) ²}
The formulae for getting the gradient :
(y₁ - y₂) /(x₁ - x₂)
CALCULATIONS :
Gradient for AB :
(0 - 1) / (9 - [-2]) = - 1 / 11
Gradient for DC = gradient for AB
Let point D = ( 1, y) then:
(6 - Y) / (4 - 1) = - 1/11
6 - y / 3 = - 1/11
11(6 - Y) = - 1 × 3
66 - 11y = - 3
- 11y = - 99
y = 9
LENGTH :
AB = DC
√(-1)² + (11)² = √122 = 11.0454 units
AB = DC = 11.0454units
DA = BC
√(6²) + (-5)² = √61 = 7.8102 units
DA = BC = 7.8102Units.
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