Show that points P(1,-2), Q(5,2), R(3,-1), S(-1,-5) are the vertices of a parallelogram
11. Show that points P(1,-2), Q(5,2), R(3,-1), S(-1,-5) are the vertices of a parallelogram
The given points are P(1, –2), Q(5, 2), R(3, –1), S(–1, –5).
\[\text { Slope of PQ } = \frac{2 + 2}{5 - 1} = \frac{4}{4} = 1\]
\[\text { Slope of QR } = \frac{- 1 - 2}{3 - 5} = \frac{- 3}{- 2} = \frac{3}{2}\]
\[\text { Slope of RS } = \frac{- 5 + 1}{- 1 - 3} = \frac{- 4}{- 4} = 1\]
\[\text { Slope of PS }= \frac{- 5 + 2}{- 1 - 1} = \frac{- 3}{- 2} = \frac{3}{2}\]
Slope of PQ = Slope of RS and Slope of QR = Slope of PS
So, PQ || RS and QR || PS
Thus, PQRS is a parallelogram.
Explanation:-
The given points are P(1, -2), Q(5, 2), R(3, -1), and S(-1, -5).
To check whether PQRS is a parallelogram or not, we need to check if its opposite sides are parallel.
First, we find the slope of each line segment:
Slope of PQ: (2 – (-2)) / (5 – 1) = 4 / 4 = 1 Slope of QR: (2 – (-1)) / (5 – 3) = 3 / 2 Slope of RS: (-5 – (-1)) / (-1 – 3) = -4 / -4 = 1 Slope of PS: (-5 – (-2)) / (-1 – 1) = -3 / -2 = 3 / 2
We observe that the slope of PQ is equal to the slope of RS, and the slope of QR is equal to the slope of PS.
Since opposite sides have the same slope, we conclude that PQRS is a parallelogram.
Therefore, we have shown that the given vertices form a parallelogram.
Chapter 5. Co-ordinate Geometry – Problem set 5 (Page 122)