Question

If (-1,2),(4,1),(7,16) are the three vertices of a parallelogram taken in order, then the fourth vertex and also the areas of the parallelogram are

• A. (-4,3), 16 sq. unit
• B. (2,17), 78 sq. unit
• C. (-8,3), 24 sq. unit
• D. (10,-5) 36 sq. unit

Answer 1

The correct option is B (2,17), 78 sq. unit

Let vertex D be (a,b)

Slope of AB=slope of DC

$\frac{1-2}{4+1}=\frac{16-b}{7-a}$

$\frac{-1}{5}=\frac{16-b}{7-a}$

$a-7=80-5b$

$a+5b=87\to i$

Slope of BC=Slope of AD

$\frac{16-1}{7-4}=\frac{b-2}{a+1}$

$\frac{15}{3}=\frac{b-2}{a+1}$

$5(a+1)=b-2$

$5a+5=b-2$

$5a-b=-7\to ii$

$ii\times 5+1$

$a+5b=87$

$25a-5b=-35$

$26a=52$

$a=2,b=5a+7=5\left(2\right)+7=17$

$\therefore 4^{th}$ vertex$=(2,17)$

Area$=\frac{1}{2}\left|\begin{array}{cccc}-1& 4& 7& 2\\ 2& 1& 16& 17\end{array}\right|$

$=\frac{1}{2}|-1-8+64-7+119-32|$

$=\frac{1}{2}|64-16+87|$

$=\frac{156}{2}=78$ sq. units