Answers

2016-03-18T20:51:30+05:30
Consider a ||gram whose vertices are A(1,-2), B(2,3), C(-3,2) and D(-4,-3).
Construction :- Draw a perpendicular(DL) to the base AB from D.
            DL is the height of the parallelogram if base is AB.
Let  D≡(p,q).
Evaluation :-
    Slope of a line = (y
 - y)/(x₂ - x)
∴ slope of line AB = (3 + 2)/(2 - 1) = 5 = m₁   (say)

slope of line AL = slope of line AB = m₁ = (q + 2)/(p - 1) 
                 (q + 2)/(p - 1) = 5
                  ⇒ q + 2 = 5p - 5
                  ⇒ 5p - q = 7 --------------------- (i)
slope of line DL = m₂ = (q + 3)/(p + 4).
We know that product of slopes of two perpendicular lines = -1
i.e, m₁ × m₂ = -1
⇒   5 × m₂ = -1
⇒        m₂ = -1/5
∵  m₂ = (q+3)/(p+4)
⇒ (q+3)/(p+4) = -1/5
⇒ -p - 4 = 5q + 15
⇒ 5q + p = -19 -----------------------(ii)
After solving the equation i) and (ii)
p = 9/13
and q = -46/13
                                ∴ L(9/13, -46/13) and D(-4, -3)
Now you can find length of DL .DL = \sqrt{ (9/13 + 4)^{2}  + (-46/13 +3)^{2} } 

DL = 61.4/13
Therefore height of ||gram = DL + 61.4/13 unit

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