(i) (b): (3, 7)
Midpoint of (2, 1) and (4, 13) is given by (3, 7).
(ii) (a): 4
\(\text { Slope }=\frac{12-0}{1-(-2)}=\frac{12}{3}=4\) 
(iii) (a): y – 2 = 4x
Equation of BC: = (y – 0)  = 4(x + 2)  
= y – 2 = 4x
(iv) (b): it will not lie on the line BC.
Equation of BC is y – 2 – 4 x = 0. Putting coordinates of A (3, 7) in this equation, we get: 7 – 2 – 12 ≠ 0, therefore we can say that point A will not lie on the line BC.
(v) (b): perpendicular
The product of the slopes is \(-2 \times \frac{1}{2}=-1\) 

(i) (c): 0
\(\text { Slope of } A B=\frac{3-3}{8-1}=0\) 
(ii) (c): \(\frac{13}{14}\) 
\(\text { Slope of } A C=\frac{16-3}{8-1}=\frac{13}{7}\) 
(iii) (a): parallel to x-axis
If slope of a line is zero, then the line is parallel to x-axis.
(iv) (d): 13x – 7y = –8
\(\text { Equation of line } \mathrm{AC}: y-3=\frac{16-3}{8-1}(x-1)\) 
⇒ 7y – 21 = 13x – 13
⇒ 13 x – 7y = – 8
(v) (a): \(\tan ^{-1}\left(\frac{m_{2}-m_{1}}{1+m_{1} m_{2}}\right)\) 
Angle between two lines, then
\(\tan \theta=\frac{m_{2}-m_{1}}{1+m_{1} m_{2}}\) 
\(\Rightarrow \theta=\tan ^{-1}\left(\frac{m_{2}-m_{1}}{1+m_{1}+m_{2}}\right)\) 

(i) (d): 90°
Angle formed from the center to the tangent of the circle is 90°.
(ii) (c):30°
Since, triangle OPT and OPRare congruent, therefore OP is angle bisector of angle TPR. Therefore, angle OPT = 30°.
(iii) (b): 2a
In triangle OPT,
\(\sin 30^{\circ}=\frac{O T}{O P}\) 
\(\Rightarrow \frac{1}{2}=\frac{a}{O P}\) 
⇒ OP = 2a
(iv) (c): \(\sqrt{3} a\) 
By using pythagorus theorem in triangle OPT, we get,
OP² = OT² + PT²
\(\Rightarrow P T=\sqrt{O P^{2}-O T^{2}}\) 
= \(\sqrt{(2 a)^{2}-a^{2}}\) 
= \(\sqrt{3} a\) 
(v) (d):  x² + y² = 4
Equation of circle: (x – 0)² + (y – 0)² = 22
⇒x² + y²  = 4

(i) (b): x + y = – 1
Coordinates which satisfy the equation of the line is x + y = –1.
(ii) (d): x² + y² – 4 x + 6 y – 51 = 0
Equation of circle : (x – 2)² + (y + 3)2 = 8²x² + y² – 4x + 6y – 51 = 0.
(iii) (d): four times
Let the radius = a, then area = \(\pi a^{2}\) If the radius be double, then area = \(\pi(2 a)^{2}=4(\pi 4 a)^{2}\). Therefore change in area is four times.
(iv) (b): \(2 \pi-4\) 
Area of square = 4 m²
Area of circle = \(\pi(\sqrt{2})^{2}=2 \pi\) 
Required area = \(2 \pi-4\) 
(v) (a): Any line perpendicular to the tangent of the circle will always pass through the centre of the circle.

(i) (c): x² = 4 ay
The equation of the parabola is of the form x² = 4 ay (as it opening upwards).
(ii) (a): \(\left(\frac{5}{2}, 10\right)\)
It can be clearly seen from the given figure that the parabola passes through point \(\left(\frac{5}{2}, 10\right)\)
(iii) (b): \(\frac{5}{32}\)
It can be clearly seen that the parabola passes through point \(\left(\frac{5}{2}, 10\right)\)
\( \left(\frac{5}{2}\right)^{2} =4 a(10) \)
\(\Rightarrow \ a =\frac{25}{4 \times 4 \times 10} \)
\(=\frac{5}{32} \)
(iv) (c): \(x^{2}=\frac{5}{8} y\)
The equation of parabola is
x² = 4ay
\(x^{2}=4\left(\frac{5}{32}\right) y=\left(\frac{5}{8}\right) y\) 
(v) (c): 2.23 m
Since, the equation of parabola is
\(x^{2}=\frac{5}{8} y\)

\( \Rightarrow \ \mathrm{AB} =2 \times \sqrt{\frac{5}{4}} \mathrm{~m} \)
\(=\sqrt{5} \mathrm{~m} \)
= 2.23 m

(i) (c): x² = 4ay
The equation of the parabola is of the form x² = 4ay(as it is upwards).
(ii) (b): (50, 24)
Here, AB = 30 m, OC = 6 m, and BC = \(=\frac{100}{2}=50 \mathrm{~m}\)
The coordinate of point A are (50, 30 – 6)
= (50, 24)
(iii) (c): \(\frac{625}{24}\)
Since A (50, 24) is a point on the parabola.
(50)² = 4a(24)
\( \Rightarrow \ a =\frac{50 \times 50}{4 \times 24} \)
\(=\frac{625}{24}\) 
(iv) (d): 6x² = 625y
Equation of the parabola,
\(x=4 \times \frac{625}{24} \times y\)
6x² = 625y
(v) (c): 9.11 m
The x-coordinate of point D is 18.
Hence, at x = 18,
6(18)² = 625y
\(\Rightarrow \ y=\frac{6 \times 18 \times 18}{625}\)
⇒ y = 3.11 (approx.)
DE = 3.11 m
DF = DE + EF
= 3.11 m + 6 m
= 9.11 m