Question

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

Answer 1

Step 1: Use the area of the triangle.

Let the given vertices be $A\left(2,1\right)$ and $B\left(3,-2\right)$ also, the third vertex be $C\left(x,y\right)$.

Apply the area of triangle that is $∆=\frac{1}{2}\left|\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]\right|$ .

Substituting the values, we get

$\begin{array}{rcl}& \Rightarrow & 5=\frac{1}{2}\left|\left[2\left(-2-y\right)+3\left(y-1\right)+x\left(1+2\right)\right]\right|\\ & \Rightarrow & 5=\frac{1}{2}\left|4-2y+3y-3+3x\right|\\ & \Rightarrow & 5=\frac{1}{2}\left|3x+y-7\right|\end{array}$

Due to modulus there will be two values first is positive and the second is negative so write the obtained equation accordingly and then form two equations from it.

$\begin{array}{rcl}5& =& \pm \frac{1}{2}\left(3x+y-7\right)\\ & \Rightarrow & \pm 10=3x+y-7\end{array}$

So, the two equations are

$3x+y=17$ or $3x+y=-3$

Step 2: Find the first set of coordinates.

Use the equation $3x+y=17$.

Substituting the given equation $y=x+3$ into $3x+y=17$, we get

$\Rightarrow \begin{array}{rcl}3x+\left(x+3\right)& =& 17\end{array}$

$\begin{array}{rcl}& \Rightarrow & 4x+4=17\end{array}$

$\begin{array}{rcl}& \Rightarrow & 4x=14\end{array}$

$\begin{array}{rcl}\therefore x& =& \frac{7}{2}\end{array}$

Substitute $x=\frac{7}{2}$ into $y=x+3$ to find the value of $y$

$\begin{array}{rcl}y& =& \frac{7}{2}+3\\ y& =& \frac{7+6}{2}\\ y& =& \frac{13}{2}\end{array}$

Step 3: Find the second set of coordinates.

Use the equation $3x+y=-3$.

Substituting the given equation $y=x+3$ into $3x+y=-3$, we get

$\Rightarrow \begin{array}{rcl}3x+\left(x+3\right)& =& -3\end{array}$

$\begin{array}{rcl}& \Rightarrow & 4x+3=-3\end{array}$

$\begin{array}{rcl}& \Rightarrow & 4x=-6\end{array}$

$\begin{array}{rcl}\therefore x& =& -\frac{3}{2}\end{array}$

Substitute $x=-\frac{3}{2}$ into $y=x+3$ to find the value of $y$

$\begin{array}{rcl}y& =& -\frac{3}{2}+3\\ y& =& \frac{-3+6}{2}\\ y& =& \frac{3}{2}\end{array}$

The obtained values of $x$ and $y$ represents two possible points $\left(\frac{7}{2},\frac{13}{2}\right)$ or $\left(-\frac{3}{2},\frac{3}{2}\right)$.

Final Answer:

Hence, the required coordinates of the third vertex is either $\left(\frac{7}{2},\frac{13}{2}\right)$ or $\left(-\frac{3}{2},\frac{3}{2}\right)$.