Question

The coordinates of points $A,B,C$ and $P$ are $(2,3),(1,2),(4,3)$ and $(t,t^2)$ respectively, where $t>0$. If the area of $△ABC$ is twice the area of $△PAB$, then the number of value(s) of $t$ is

Answer 1

Given coordinates are $A=(2,3),B=(1,2),C=(4,3)$ and $P=(t,t^2)$

Area of $△ABC$
$=\frac{1}{2}\left|\begin{array}{cccc}2& 1& 4& 2\\ 3& 2& 3& 3\end{array}\right|=\frac{1}{2}|1-5+6|=1$

Area of $△PAB$
$=\frac{1}{2}\left|\begin{array}{cccc}t& 2& 1& t\\ t^2& 3& 2& t^2\end{array}\right|=\frac{1}{2}|3t-2t^2+1+t^2-2t|=\frac{1}{2}|-t^2+t+1|$

Given, the area of $△ABC$ is twice that of $△PAB$, so
$1=2\times \frac{1}{2}|-t^2+t+1|\Rightarrow -t^2+t+1=\pm 1\Rightarrow t^2-t-1=\pm 1\Rightarrow t^2-t=0,2$

Now, $t^2-t=0$
$\Rightarrow t=0,1$

Also, $t^2-t-2=0$
$\Rightarrow (t-2)(t+1)=0\Rightarrow t=-1,2$

Hence, $t=1,2$ as $t>0$