Question

List-II gives distance between pair of points given in List-I. Match them correctly.
$\begin{array}{cccc}& List-I& & ListII\\ \left(P\right)& (-5,7),(-1,3)& \left(1\right)& 5\\ \left(Q\right)& (5,6),(1,3)& \left(2\right)& \sqrt{8}\\ \left(R\right)& (\sqrt{3}+1,1),(0,\sqrt{3})& \left(3\right)& \sqrt{6}\\ \left(S\right)& (0,0),(-\sqrt{3},\sqrt{3})& \left(4\right)& 4\sqrt{2}\end{array}$

• A. P-4, Q-3, R-2, S-1
• B. P-4, Q-1, R-2, S-3
• C. P-3, Q-2, R-4, S-1
• D. P-1, Q-2, R-3, S-4

Answer 1

The correct option is B

P-4, Q-1, R-2, S-3

(b): $P\to 4;Q\to 1;R\to 2;S\to 3$

(P) $(-5,7),(-1,3)=\sqrt{(-1+5{)}^2+(3-7{)}^2}=\sqrt{16+16}=4\sqrt{2}$

(Q) $(5,6),(1,3)=\sqrt{(1-5{)}^2+(3-6{)}^2}=\sqrt{16+9}=5$

(R) $(\sqrt{3}+1,1),(0,\sqrt{3})$
$=\sqrt{(0-(\sqrt{3}+1){)}^2+(\sqrt{3}-1{)}^2}$
$=\sqrt{(\sqrt{3}+1{)}^2+(\sqrt{3}-1{)}^2}$
$=\sqrt{3+2\sqrt{3}+1+3-2\sqrt{3}+1}$
$=\sqrt{6+2}=\sqrt{8}=2\sqrt{2}$

(S) $(0,0),(-\sqrt{3},\sqrt{3})=\sqrt{(-\sqrt{3}{)}^2+(\sqrt{3}{)}^2}=\sqrt{3+3}=\sqrt{6}$