# If the points A(1,-2) , B(2,3) , C(-3,2) and D(-4,-3) are the vertices of parallelogram ABCD ,then taking AB as base , find the height of the parallelogram .

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by pradhi

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by pradhi

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DL is the height of the parallelogram if base is AB.

Let D≡(p,q).

Slope of a line = (y

∴ 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 = 61.4/13

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