Question

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

• A. $(\frac{7}{2},\frac{13}{2})$
• B. $(\frac{5}{2},\frac{5}{2})$
• C. $(-\frac{3}{2},\frac{3}{2})$
• D. $(0,0)$

Answer 1

Answer: (A), (C)

$(-\frac{3}{2},\frac{3}{2})$
$(\frac{7}{2},\frac{13}{2})$

Given that the vertices of the $\Delta ABC$ are $A=(x_1,y_1)=(2,1),B=(x_2,y_2)=(3,-2)$ and take $C=(p,q)$.

$\because$ $C$ lies on the line, $y=x+3\Rightarrow q=p+3$ ..........(i)

Now, $\Delta =\pm 5$ ...($\because$ the $\Delta$ may lie on the either side of the axes.)

$\therefore ar\Delta ABC=\frac{1}{2}\left|\begin{array}{ccc}p& q& 1\\ x_1& y_1& 1\\ x_2& y_2& 1\end{array}\right|=\frac{1}{2}\left|\begin{array}{ccc}p& q& 1\\ 2& 1& 1\\ 3& -2& 1\end{array}\right|=\pm 5$

$\Rightarrow q+3p-7=\pm 10.$

Either $q+3p-7=10$

$\Rightarrow q=17-3p$ ...........(ii)

Or $q+3p-7=-10$

$\Rightarrow q=-3p-3$ ......(iii)

Eliminating $q$ from equations (i) & (ii), we have

$p+3=17-3p$

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

Putting this value of $p$ in equation (i), we get

$q=\frac{13}{2}$

$\therefore (p,q)=(\frac{7}{2},\frac{13}{2}).\to$ First set

Or eliminating $q$ from equations (i) & (iii), we have

$p+3=-3p-3$

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

Putting this value of $p$ in equation (i), we get

$q=-\frac{3}{2}$

$\therefore (p,q)=(-\frac{3}{2},\frac{3}{2})\to$ Second set