Four vertices of a quadrilateral are located at $(7,6),(-5,1),(-2,-3)$, and $(10,2)$. The area of the quadrilateral in square units is
(A) 60
(B) 63
(C) 67
(D) 70
(E) 72
We begin by constructing rectangle $A B C D$ around the given quadrilateral $P Q R S$, as shown.

The vertical sides $A B$ and $D C$ pass through points $Q$ and $S$, respectively.

The horizontal sides $A D$ and $B C$ pass through points $P$ and $R$, respectively.

We determine the area of $P Q R S$ by subtracting the areas of the four right-angled triangles, $A Q P, Q B R, C S R$, and $S D P$, from the area of $A B C D$.

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-135.jpg?height=393&width=596&top_left_y=1229&top_left_x=1358)

To determine the horizontal side lengths of the right-angled triangles we count units along the $x$-axis, or we subtract the $x$-coordinates of two vertices.

For example, since $A B$ is vertical and passes through $Q(-5,1)$, the $x$-coordinates of $A$ and $B$ are equal to that of $Q$, which is -5 .

Thus, the length of $A P$ is determined by subtracting the $x$-coodinate of $A$ from the $x$-coordinate of $P$, which is 7 .

Therefore the length of $A P$ is $7-(-5)=12$.

Similarly, the length of $B R$ is $-2-(-5)=3$.

Since $D C$ is vertical and passes through $S(10,2)$, the $x$-coordinates of $D$ and $C$ are equal to that of $S$, which is 10 .

Thus, the length of $P D$ is $10-7=3$, and the length of $R C$ is $10-(-2)=12$.

To determine the vertical side lengths of the right-angled triangles we may count units along the $y$-axis, or we may subtract the $y$-coordinates of two vertices.

For example, since $A D$ is horizontal and passes through $P(7,6)$, the $y$-coordinates of $A$ and $D$ are equal to that of $P$, which is 6 .

Thus, the length of $A Q$ is determined by subtracting the $y$-coodinate of $Q$ (which is 1 ) from the $y$-coordinate of $A$.

Therefore the length of $A Q$ is $6-1=5$.

Similarly, the length of $D S$ is $6-2=4$.

Since $B C$ is horizontal and passes through $R(-2,-3)$, the $y$-coordinates of $B$ and $C$ are equal to that of $R$, which is -3 . Thus, the length of $Q B$ is $1-(-3)=4$, and the length of $S C$ is $2-(-3)=5$.

The area of $\triangle A Q P$ is $\frac{1}{2} \times A Q \times A P=\frac{1}{2} \times 5 \times 12=30$.

The area of $\triangle C S R$ is also 30 .

The area of $\triangle Q B R$ is $\frac{1}{2} \times Q B \times B R=\frac{1}{2} \times 4 \times 3=6$.

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-136.jpg?height=393&width=488&top_left_y=150&top_left_x=1431)

The area of $\triangle S D P$ is also 6 .

Since $A B=A Q+Q B=5+4=9$ and $B C=B R+R C=3+12=15$, the area of $A B C D$ is $9 \times 15=135$.

Finally, the area of $P Q R S$ is $135-30 \times 2-6 \times 2=135-60-12=63$.

ANSWER: (B)

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