(i) (d): Since the coordinates of A and Bare (2, 3) and (2, 1) respectively.
\(\therefore\) Required distance = AB
\(=\sqrt{(2-2)^{2}+(1-3)^{2}}=\sqrt{2^{2}}=2 \text { units }\)
(ii) (b): Since, library is situated at C(4, 1)
\(\therefore\)Required distance = BC
\(=\sqrt{(4-2)^{2}+(1-1)^{2}}=\sqrt{2^{2}+0^{2}}=2 \text { units }\)
(iii) (d): Required distance = AC
\(=\sqrt{(4-2)^{2}+(1-3)^{2}}=\sqrt{2^{2}+2^{2}}=2 \sqrt{2} \text { units }\)
(iv) (b): Since AB = BC \(\neq\) AC, therefore \(\Delta\)ABC is an isosceles triangle.
(v) (d): Distance between O and A
\(=\sqrt{2^{2}+3^{2}}=\sqrt{4+9}=\sqrt{13} \text { units }\)
and distance between O and B =
\(\sqrt{2^{2}+1^{2}}=\sqrt{4+1}=\sqrt{5} \text { units }\)
Thus, required distance = \((\sqrt{13}-\sqrt{5}) \text { units }\)

Consider the house is at origin (0, 0), then coordinates of grocery store, electrician's shop, food cart and bus stand are respectively (2, 3), (-4, -6), (6, - 8) and (-6, 8)
(i) (d): Since, grocery store is at (2, 3) and food cart is at (6, -8)
\(\therefore\) Required distance \(=\sqrt{(6-2)^{2}+(-8-3)^{2}}\)
\(=\sqrt{4^{2}+11^{2}}=\sqrt{16+121}=\sqrt{137} \mathrm{~cm}\)
(ii) (b): Required distance
\(=\sqrt{(-6)^{2}+8^{2}}=\sqrt{36+64}=\sqrt{100}=10 \mathrm{~cm}\)
(iii) (c) : Let 0divides EG in the ratio k : I,then

\(\begin{array}{l} 0=\frac{2 k-4}{k+1} \\ \Rightarrow \quad 2 k=4 \\ \Rightarrow k=2 \end{array}\)
Thus, 0 divides EG in the ratio 2 : 1
Hence, required ratio = OG :OE i.e., 1 : 2
(iv) (c): Since, (0, 0) is the mid-point of (-6,8) and (6, -8), therefore both bus stand and food cart are at equal distances from the house. Hence, required ratio is 1 : 1.
(v) (d): Mid-point of grocery store and electrician's shop is \(\left(\frac{2-4}{2}, \frac{3-6}{2}\right), \text { i.e., }\left(-1, \frac{-3}{2}\right)\) Thus, the diagonals does not bisect each other [\(\because\) Mid-point are not same]
Hence, they form a quadrilateral.

(i) (c): We have, P( -3,4), Q(3, 4) and R( -2, -1).
\(\therefore\) Coordinates of centroid of \(\Delta\)PQR
\(=\left(\frac{-3+3-2}{3}, \frac{4+4-1}{3}\right)=\left(\frac{-2}{3}, \frac{7}{3}\right)\)
(ii) (d): Coordinates of S= \(\left(\frac{-3+3}{2}, \frac{4+4}{2}\right)=(0,4)\)
(iii) (c): Coordinates of T= \(\left(\frac{-2+3}{2}, \frac{-1+4}{2}\right)=\left(\frac{1}{2}, \frac{3}{2}\right)\)
(iv) (a): Coordinates of U = \(\left(\frac{-2-3}{2}, \frac{-1+4}{2}\right)=\left(\frac{-5}{2}, \frac{3}{2}\right)\)
(v) (c): The centroid of triangle formed by joining the mid-points of sides of given triangle is same as that of the given trangle.
So, centroid of \(\Delta S T U=\left(\frac{-2}{3}, \frac{7}{3}\right)\)

Coordinates of A, Band Care (-2, -3), (2, 3) and (3,2).
(i) (d): Required distance \(=\sqrt{(2+2)^{2}+(3+3)^{2}}\)
\(=\sqrt{4^{2}+6^{2}}=\sqrt{16+36}=2 \sqrt{13} \mathrm{~km} \approx 7.2 \mathrm{~km}\)
(ii) (d): Required distance \(=\sqrt{(3+2)^{2}+(2+3)^{2}}\)
\(=\sqrt{5^{2}+5^{2}}=5 \sqrt{2} \mathrm{~km}\)
(iii) (b): Distance between Band C
\(=\sqrt{(3-2)^{2}+(2-3)^{2}}=\sqrt{1+1}=\sqrt{2} \mathrm{~km}\)
Thus, distance travelled by first bus to reach to B
\(=A C+C B=5 \sqrt{2}+\sqrt{2}=6 \sqrt{2} \mathrm{~km} \approx 8.48 \mathrm{~km}\)
and distance travelled by second bus to reach to B
\(=A B=2 \sqrt{13} \mathrm{~km} \approx 7.2 \mathrm{~km}\)
\(\therefore\) Distance of first bus is greater than distance of the
second bus, therefore second bus should be chosen.
(iv) (d): Distance travelled by first bus = 8.48 km
\(\therefore\) Total fare = 8.48 x 10 = Rs 84.80
(v) (b): Distance travelled by second bus = 7. 2 km
\(\therefore\) Total fare = 7.2 x 15 = Rs 108
 

(i) (b): The distance between A and C
\(=\sqrt{(8-4)^{2}+(5+3)^{2}}=\sqrt{4^{2}+8^{2}} \)
\(=\sqrt{16+64}=\sqrt{80}=4 \sqrt{5} \text { units }\)
(ii) (a): Let the coordinates of I be (x, y).

Then, by section formula
\(x =\frac{1 \times 8+2 \times 7}{1+2}=\frac{8+14}{3}=\frac{22}{3}\)
\(\text { and } y =\frac{1 \times 5+2 \times 3}{1+2}=\frac{5+6}{3}=\frac{11}{3}\)
Thus, the coordinates of I is \(\left(\frac{22}{3}, \frac{11}{3}\right)\)
(iii) (b): The mid -point of A and C
\(=\left(\frac{8+4}{2}, \frac{5-3}{2}\right)=(6,1)\)
(iv) (b): Let B divides the line segment joining A and C in the ratio k : 1. Then, the coordinates of B will be
\(\left(\frac{8 k+4}{k+1}, \frac{5 k-3}{k+1}\right)\)
\(\text { Thus, we have }\left(\frac{8 k+4}{k+1}, \frac{5 k-3}{k+1}\right)=(7,3)\)
\(\Rightarrow \frac{8 k+4}{k+1}=7 \text { and } \frac{5 k-3}{k+1}=3 \)
\(\text { Consider, } \frac{8 k+4}{k+1}=7 \Rightarrow 8 k+4=7 k+7 \Rightarrow k=3\)
Hence, the required ratio is 3 : 1
(v) (a):\(\because\) B divides AC in the ratio 3 : 4.
\(\therefore\) A, B, C lie on a straight line.