Question

Let the coordinates of $P,Q$ and $R$ be $(6,3),$ $(-3,5)$ and $(4,-2)$ respectively. If $A\equiv (x,y),$ then the ratio of area of $△AQR$ to $△PQR$ is

• A. $\frac{2}{7}$ if $x=0,y=0$
• B. $\frac{2}{7}$ if $x=1,y=2$
• C. $1$ if $x=5,y=4$
• D. $1$ if $x=4,y=5$

Answer 1

Answer: (A), (C), (D)

$1$ if $x=4,y=5$

Given : $P=(6,3),Q=(-3,5),R=(4,-2)$
Now, $Areaof\Delta PQR$
$=\frac{1}{2}\left|\begin{array}{cccc}{x}_1& x_2& x_3& x_1\\ y_1& y_2& y_3& y_1\end{array}\right|=\frac{1}{2}\left|\begin{array}{cccc}6& -3& 4& 6\\ 3& 5& -2& 3\end{array}\right|=\frac{1}{2}\left|\right(30+9)+(6-20)+(12+12\left)\right|=\frac{49}{2}$

Also, $Areaof\Delta AQR$
$=\frac{1}{2}\left|\begin{array}{cccc}x& -3& 4& x\\ y& 5& -2& y\end{array}\right|=\frac{1}{2}\left|\right(5x+3y)+(6-20)+(4y+2x\left)\right|=\frac{|7x+7y-14|}{2}$

Now, $\frac{Areaof\Delta AQR}{Areaof\Delta PQR}=\frac{|x+y-2|}{7}$

Therefore, when $x=0,y=0$, we get
$\frac{Areaof\Delta AQR}{Areaof\Delta PQR}=\frac{2}{7}$

When $x=1,y=2$, we get
$\frac{Areaof\Delta AQR}{Areaof\Delta PQR}=\frac{1}{7}$

When $x=5,y=4$, we get
$\frac{Areaof\Delta AQR}{Areaof\Delta PQR}=1$

When $x=4,y=5$, we get
$\frac{Areaof\Delta AQR}{Areaof\Delta PQR}=1$