Question

The area of a triangle is 5. Two of its vertices are (2, 1) and (3, -2). The third vetex lies on y = x + 3. Find the third vertex:

• A. $(\frac{2}{7},\frac{13}{5})$
• B. $(\frac{7}{2},\frac{13}{2})$
• C. $(\frac{9}{2},\frac{13}{2})$
• D. $(\frac{7}{2},\frac{13}{2})or(-\frac{3}{2},\frac{3}{2})$

Answer 1

The correct option is D $(\frac{7}{2},\frac{13}{2})or(-\frac{3}{2},\frac{3}{2})$

Let $x_1,y_1$ be the third vetex then $y_1=x_1+3\left(1\right)$
The area of the triangle formed by the points (2, 1) (3, -2) and $(x_1,y_1)$
$=\frac{1}{2}\left(-4-3+3y_1+2x_1+x_1-2y_1\right)=\frac{1}{2}(3x_1+y_1-7)$
By the given condition.
$\pm \frac{1}{2}(3x_1+y_1-7)=5\Rightarrow 3x_1+y_1=\pm 10\Rightarrow 3x_1+y_1=\mathrm{17.......}\left(2\right)and3x_1+y_1=-\mathrm{3........}\left(3\right)$
$Solving\left(1\right)and\left(2\right)weget{x}_1=\frac{7}{2},y_1=\frac{13}{2}Solving\left(1\right)and\left(3\right)weget{x}_1=-\frac{3}{2}and{y}_1=\frac{3}{2}$
Hence, the coordinates of the third vertex is either $(\frac{7}{2},\frac{13}{2})or(-\frac{3}{2},\frac{3}{2})$