Question

$O(0,0),A(1,1),B(0,3)$ are the vertices of $△OAB,P$ divides $OB$ in the ratio $1:2,Q$ is the mid-point of $AP,R$ divides $AB$ in the ratio $2:1$.

If $\alpha (QR{)}^2=\beta (PR{)}^2$, then

• A. $\alpha -\beta =3$
• B. $\alpha -\beta =-3$
• C. $\alpha =\beta$
• D. $\alpha =3\beta$

Answer 1

Answer: (A)

$\alpha -\beta =3$

$P$ divides $OB$ in $1:2$

$P\equiv (\frac{2\times 0+1\times 0}{2+1},\frac{2\times 0+1\times 3}{2+1})\Rightarrow P\equiv (0,1)$

$Q$ is mid point of $AP$
$\Rightarrow Q=(\frac{0+1}{2},\frac{1+1}{2})\equiv (\frac{1}{2},1)$

$R$ divides $AB$ in $2:1$

$R\equiv (\frac{2\times 0+1\times 1}{2+1},\frac{2\times 3+1\times 1}{2+1})\equiv (\frac{1}{3},\frac{7}{3})$

$\therefore {\left(QR\right)}^2={(\frac{1}{3}-\frac{1}{2})}^2+{(\frac{7}{3}-1)}^2=\frac{1}{36}+\frac{16}{9}=\frac{65}{36}$

${\left(PR\right)}^2={(\frac{1}{3}-0)}^2+{(\frac{7}{3}-1)}^2=\frac{1}{9}+\frac{16}{9}=\frac{17}{9}$

$\therefore \alpha -\beta =3$