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 third vertex.

Answer 1

The given vertices are $(2,1)$ and $(3,-2)$ and the area of triangle is $5$ sq. unit

Let the third vertex be $(x_3,y_3)$

Area of triangle $=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$
Here, $x_1=2,y_1=1;x_2=3,y_2=-2$; area of the triangle$=5$ sq unit
$\Rightarrow 5=\frac{1}{2}|2(-2-y_3)+3(y_3-1)+x_3(1+2)|$

$\Rightarrow 10=|-4-2y_3+3y_3-3+3x_3|$
$\Rightarrow 10=|3x_3+y_3-7|$
$\Rightarrow 3x_3+y_3-7=\pm 10$

Taking positive sign
$\Rightarrow 3x_3+y_3-7=10$

$\Rightarrow 3x_3+y_3=17$ .....(i)

Taking negative sign
$\Rightarrow 3x_3+y_3-7=-10$
$\Rightarrow 3x_3+y_3=-3$ .......(ii)

Given that $(x_3,y_3)$ lies on $y=x+3$
So $y_3=x_3+3$

$\Rightarrow x_3-y_3=-3$ ........(iii)

For (i) and (iii)

Adding eq (i) & (iii)

$4x_3=14$

$\Rightarrow x_3=\frac{7}{2}$

Putting in (iii)

$\Rightarrow \frac{7}{2}-y_3=-3$

$\Rightarrow y_3=\frac{7}{2}+3$

$\Rightarrow y_3=\frac{13}{2}$

$x_3=\frac{7}{2},y_3=\frac{13}{2}$

For eq (ii) and (iii)

Adding eq (ii) & (iii)

$4x_3=-6$

$\Rightarrow x_3=\frac{-6}{4}$

$\Rightarrow x_3=\frac{-3}{2}$

Putting in (iii)

$\Rightarrow -\frac{3}{2}-y_3=-3$

$\Rightarrow y_3=-\frac{3}{2}+3$

$\Rightarrow y_3=\frac{3}{2}$

$x_3=\frac{-3}{2},y_3=\frac{3}{2}$

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