Question

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 coordinates of the third vertex.

Answer 1

Let the third vertex be A $(x,y)$. Other two vertices of the triangle are B (2, 1) and C (3, -2).

We have ,

$\therefore$ Area of $△$ABC = 5 sq.units

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

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

$\Rightarrow$ $\frac{1}{2}[3x+y-7]=5$

Now, either

$\Rightarrow$ $3x+y-17=0$ or $3x+y+3=0$

It is given that the vertex A $(x,y)$ lies on $y=x+3$

Solving $3x+y-17=0$ and $y=x+3$, we get $x=\frac{7}{2}andy=\frac{13}{2}$

Solving $3x+y+3=0$ and $y=x+3$, we get $x=\frac{-3}{2}andy=\frac{3}{2}$

Hence, the coordinates of the third vertex are $(\frac{7}{2},\frac{13}{2})or(\frac{-3}{2},\frac{3}{2})$