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

Given:

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

To do:

We have to find the third vertex.

Solution:

Let $A(2, 1), B(3, -2)$ and C(x, y)$ be the vertices of $\triangle ABC$.

We know that,

Area of a triangle with vertices $(x_1,y_1), (x_2,y_2), (x_3,y_3)$ is given by, 

Area of $\Delta=\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$

Therefore,

Area of triangle \( ABC=\frac{1}{2}[2(-2-y)+3(y-1)+x(1+2)] \)

\( 5=\frac{1}{2}[-4-2y+3y-3+3x] \)

\( 5(2)=(y-7+3x) \)

\( 10+7=3x+y \)

\( 3x+y=17 \)

The third vertex lies on $y = x + 3$.

This implies,

\( y=x+3 \)

\( \Rightarrow 3x+(x+3)=17 \)

\( \Rightarrow 4x=17-3 \)

\( \Rightarrow 4x=14 \)

\( \Rightarrow x=\frac{14}{4} \)

\( \Rightarrow x=\frac{7}{2} \)

\( \Rightarrow y=\frac{7}{2}+3 \)

\( \Rightarrow y=\frac{7+2(3)}{2} \)

\( \Rightarrow y=\frac{13}{2} \)

The third vertex is $(\frac{7}{2}, \frac{13}{2})$.