The points A ( 1, -2 ), B ( 2, 3 ), C ( k, 2 ) and D ( - 4, - 3 ) are the vertices of a parallelogram. Find the value of k and the altitude of the parallelogram corresponding to the base AB.
As, In a parallelogram opposite sides are equal
Therefore,
AB = CD
And we know, distance formula, i.e. distance between two points P(x1, y1) and Q(x2, y2)
Using this we get,
√(2 - 1)2 + (3 - (-2))2 = √(-4 - k)2 + (-3 - 2)2
Squaring both side, and solving
⇒ 1 + 25 = 16 + k2 + 8k + 25
⇒ k2 + 8k + 15 = 0
⇒ k2 + 5k + 3k + 15 = 0
⇒ k(k + 5) + 3(k + 5) = 0
⇒ (k + 3)(k + 5) = 0
⇒ k = -3 and k = -5 …[1]
Also,
AD = BC [opposite sides should be equal]
Using distance formula, we get,
√(-4 - 1)2 + (-3 - (-2))2 = √k - 2)2 + (2 - 3)2
Squaring both side, and solving
⇒ 25 + 1 = k2 - 4k + 4 + 1
⇒ k2 - 4k - 21 = 0
⇒ k2 - 7k + 3k - 21 = 0
⇒ k(k - 7) + 3(k - 7) = 0
⇒ (k + 3)(k - 7) = 0
⇒ k = -3 and k = 7 …[2]
From [1] and [2]
k = -3
Also,
By distance formula, length of AB is,
AB = √(2 - 1)2 + (3 - (-2))2
⇒ AB = √(1 + 25) = √26 units
Area of parallelogram = 2 × Area of ABD Δ
we know area of triangle formed by three points (x1, y1), (x2,y2) and (x3, y3)
⇒ area of parallelogram = 2(12) = 24 sq. units
Also,
Area of parallelogram ABCD = Base AB × Altitude of Parallelogram
⇒ 24 = √26 × Altitude
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