Question

The area of a triangle is $5$. Two of its vertices are $(2,1)$ and $(3,-2)$. The third vertex satisfies $y=x+3$. The coordinates of the third vertex can be

• A. $(-\frac{3}{2},\frac{3}{2})$
• B. $(\frac{3}{2},-\frac{3}{2})$
• C. $(\frac{7}{2},\frac{13}{2})$
• D. $(-\frac{1}{4},\frac{11}{4})$

Answer 1

Answer: (A), (C)

$(-\frac{3}{2},\frac{3}{2})$
$(\frac{7}{2},\frac{13}{2})$

As the third vertex satisfies $y=x+3$, its coordinates are of the form $(x,x+3)$.

The area of the triangle
whose vertices are $(2,1),(3,-2)$ and $(x,x+3)$ is

$\frac{1}{2}\left|\begin{array}{ccc}x& x+3& 1\\ 2& 1& 1\\ 3& -2& 1\end{array}\right|=\frac{1}{2}|4x-4|=|2x-2|$

According to the given condition, $2x-2=\pm 5$, that is, $x=-3/2,7/2$.

Thus the coordinates of the third
vertex can be $(-\frac{3}{2},\frac{3}{2})$ or $(\frac{7}{2},\frac{13}{2})$