Q. 51
Question
In Exercises 50–53 the coordinates of points P, Q, R, and S are given. (a) Determine whether quadrilateral PQRS is a parallelogram. (b) Find the area of quadrilateral PQRS.
P(−1, 3), Q(2, 5), R(4, 1), S(1, −1).
Step-by-Step Solution
VerifiedPart (a) The quadrilateral PQRS is a parallelogram.
Part (b) The area of quadrilateral PQRS is 16 square units.
A quadrilateral is a closed shape and a four-sided polygon that has four edges, four vertices, and four angles.
A parallelogram is defined as a quadrilateral with both pairs of opposite sides parallel and equal.
To find that quadrilateral PQRS is a parallelogram. We have to find the vectors of opposite sides because a quadrilateral is said to be a parallelogram if the opposite sides of a quadrilateral are equal and parallel.
So,
\(\underset{PQ}{\rightarrow}=\left< 2-(-1),5-3 \right>\)
\(\underset{PQ}{\rightarrow}=\left<3,2 \right>\)
And
\(\underset{QR}{\rightarrow}=\left< 4-2,1`-5 \right>\)
\(\underset{QR}{\rightarrow}=\left<2,-4 \right>\)
And
\(\underset{RS}{\rightarrow}=\left< 1-4,(-1)-1 \right>\)
\(\underset{RS}{\rightarrow}=\left<-3,-2 \right>\)
And
\(\underset{SP}{\rightarrow}=\left< (-1)-1,3-(-1) \right>\)
\(\underset{SP}{\rightarrow}=\left<-2,4 \right>\)
The opposite vectors of a quadrilateral are scalar multiples of each other which means opposite sides of a quadrilateral are parallel.
Now, let's find the magnitude of the vectors
\(\underset{PQ}{\rightarrow}=\left<3,2 \right>\)
\(\|\overrightarrow{PQ}\|=\sqrt{2^2+3^2}\)
\(\|\overrightarrow{PQ}\|=\sqrt{4+9}\)
\(\|\overrightarrow{PQ}\|=\sqrt{13} \quad\)
And
\(\underset{QR}{\rightarrow}=\left<2,-4 \right>\)
\(\|\overrightarrow{QR}\|=\sqrt{2^2+(-4)^2}\)
\(\|\overrightarrow{QR}\|=\sqrt{4+16}\)
\(\|\overrightarrow{QR}\|=\sqrt{20} \quad\)
And
\(\underset{RS}{\rightarrow}=\left<-3,-2 \right>\)
\(\|\overrightarrow{RS}\|=\sqrt{(-3)^2+(-2)^2}\)
\(\|\overrightarrow{RS}\|=\sqrt{9+4}\)
\(\|\overrightarrow{RS}\|=\sqrt{13} \quad\)
And
\(\underset{SP}{\rightarrow}=\left<-2,4 \right>\)
\(\|\overrightarrow{SP}\|=\sqrt{(-2)^2+(4)^2}\)
\(\|\overrightarrow{SP}\|=\sqrt{4+16}\)
\(\|\overrightarrow{SP}\|=\sqrt{20} \quad\)
Thus, the opposite sides of quadrilateral PQRS are equal in length. Hence the quadrilateral PQRS is a parallelogram because opposite sides are equal and parallel.
To find the area of quadrilateral PQRS, we will have to find the area of the parallelogram.
\(Area=\|\overrightarrow{PQ} \times \overrightarrow{Q R}\|\)
\(Area=\|\langle 3,2\rangle \times\langle 2,-4\rangle\|\)
\(Area=\|\langle 0,0,16\rangle\|\)
\(Area=\sqrt{0+0+16^2}\)
\(Area=16\)
Hence, the area of quadrilateral PQRS is 16 square units.