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P (4, 2) and Q (-1, 5) are the vertices of parallelogram PQRS and (-3, 2) are the co-ordinates of the point of intersection of the diagonals. Find co-ordinates of R and S.

Section Formula

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Answer

We know that diagonals of a parallelogram bisect each other.

Let co-ordinates of R= (a, b) and S = (c, d).

From figure,

P (4, 2) and Q (-1, 5) are the vertices of parallelogram PQRS and (-3, 2) are the co-ordinates of the point of intersection of the diagonals. Find co-ordinates of R and S. Section and Mid-Point Formula, Concise Mathematics Solutions ICSE Class 10.

O is the mid-point of PR.

By formula,

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

Substituting value we get,

⇒(−3,2)=(4+a2,2+b2)⇒−3=4+a2 and 2=2+b2⇒4+a=−6 and 2+b=4⇒a=−10 and b=2.\Rightarrow (-3, 2) = \Big(\dfrac{4 + a}{2}, \dfrac{2 + b}{2}\Big) \\[1em] \Rightarrow -3 = \dfrac{4 + a}{2} \text{ and } 2 = \dfrac{2 + b}{2} \\[1em] \Rightarrow 4 + a = -6 \text{ and } 2 + b = 4 \\[1em] \Rightarrow a = -10 \text{ and } b = 2.

R = (a, b) = (-10, 2).

O is also the mid-point of QS,

⇒(−3,2)=(−1+c2,5+d2)⇒−3=−1+c2 and 2=5+d2⇒−1+c=−6 and 5+d=4⇒c=−5 and d=−1.\Rightarrow (-3, 2) = \Big(\dfrac{-1 + c}{2}, \dfrac{5 + d}{2}\Big) \\[1em] \Rightarrow -3 = \dfrac{-1 + c}{2} \text{ and } 2 = \dfrac{5 + d}{2} \\[1em] \Rightarrow -1 + c = -6 \text{ and } 5 + d = 4 \\[1em] \Rightarrow c = -5 \text{ and } d = -1.

S = (c, d) = (-5, -1).

Hence, the co-ordinates of R = (-10, 2) and S = (-5, -1).

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