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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

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 = (y₂ - y₁)/(x₂ - x₁) ∴ 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