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In parallelogram ABCD, A = (6, 0), B = (12, -4) and C = (4, -4); then the co-ordinates of vertex D are :

  1. (2, 0)

  2. (-2, 0)

  3. (0, 2)

  4. (0, -2)

Section Formula

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Answer

Let co-ordinates of vertex D be (x, y).

We know that,

Diagonal of parallelogram bisect each other.

From figure,

In parallelogram ABCD, A = (6, 0), B = (12, -4) and C = (4, -4); then the co-ordinates of vertex D are : Section Formula and Mid-Point Formula, Concise Mathematics Solutions ICSE Class 10.

O (a, b) is the mid-point of AC.

By mid-point formula,

Mid-point = (x1+x22,y1+y22)\Big(\dfrac{x1 + x2}{2}, \dfrac{y1 + y2}{2}\Big)

Substituting values we get :

⇒(a,b)=(6+42,0+(−4)2)=(102,−42)=(5,−2).\Rightarrow (a, b) = \Big(\dfrac{6 + 4}{2}, \dfrac{0 + (-4)}{2}\Big) \\[1em] = \Big(\dfrac{10}{2}, \dfrac{-4}{2}\Big) \\[1em] = (5, -2).

From figure,

O is also the mid-point of BD.

∴(5,−2)=(12+x2,−4+y2)⇒12+x2=5 and −4+y2=−2⇒12+x=10 and −4+y=−4⇒x=10−12 and y=−4+4⇒x=−2 and y=0.\therefore (5, -2) = \Big(\dfrac{12 + x}{2}, \dfrac{-4 + y}{2}\Big) \\[1em] \Rightarrow \dfrac{12 + x}{2} = 5 \text{ and } \dfrac{-4 + y}{2} = -2 \\[1em] \Rightarrow 12 + x = 10 \text{ and } -4 + y = -4 \\[1em] \Rightarrow x = 10 - 12 \text{ and } y = -4 + 4 \\[1em] \Rightarrow x = -2 \text{ and } y = 0.

D = (-2, 0).

Hence, Option 2 is the correct option.

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