Students often turn to ML Aggarwal Class 12 Solutions ISC Case Study Based Questions Section B to clarify doubts and improve problem-solving skills.

## ML Aggarwal Class 12 Maths Solutions Case Study Based Questions Section B

Chapter 1 Vectors
Case-Study Based Questions (Solved)

Case-Study 1 :

A farmer moves along the boundary of a triangular field ABC. Three vertices of the triangular field are A (1, 2, 3), B (- 1, 0, 0) and C (0, 1, 2) respectively.

On the basis of above information, answer the following questions :

(i) The magnitude of $$\overrightarrow{\mathbf{B C}}$$ is
(a) √3
(b) √6
(c) √17
(d) 3
Solution:
(b) √6

$$\overrightarrow{\mathbf{B C}}$$ = P.V. of C – P.V. of B
= $$(0 \hat{i}+\hat{j}+2 \hat{k})-(-\hat{i}+0 \hat{j}+0 \hat{k})$$
= $$\hat{i}+\hat{j}+2 \hat{k}$$
∴ $$|\overrightarrow{\mathrm{BC}}|=\sqrt{1^2+1^2+2^2}=\sqrt{6}$$

(ii) The ∠ABC is
(a) cos-1 $$\left(\frac{10}{\sqrt{103}}\right)$$
(b) cos-1 $$\left(\frac{10}{\sqrt{105}}\right)$$
(c) cos-1 $$\left(\frac{10}{\sqrt{102}}\right)$$
(d) cos-1 $$\left(\frac{10}{\sqrt{107}}\right)$$
Solution:
(c) cos-1 $$\left(\frac{10}{\sqrt{102}}\right)$$

(iii) The projection of $$\overrightarrow{\mathbf{B A}}$$ on $$\overrightarrow{\mathbf{B C}}$$ is
(a) $$\frac{10}{\sqrt{6}}$$
(b) $$\frac{10}{\sqrt{17}}$$
(c) $$\frac{10}{\sqrt{13}}$$
(d) None of these
Solution:
(a) $$\frac{10}{\sqrt{6}}$$

The projection of $$\overrightarrow{\mathbf{B A}}$$ on $$\overrightarrow{\mathbf{B C}}$$ = $$\frac{\overrightarrow{\mathrm{BA}} \cdot \overrightarrow{\mathrm{BC}}}{|\overrightarrow{\mathrm{BC}}|}$$
= $$\frac{10}{\sqrt{6}}$$

(iv) The area of ∆ABC is
(a) $$\frac{1}{2}$$ sq. units
(b) $$[\frac{1}{\sqrt{2}}/latex] sq. units (c) 2 sq. units (d) 1 sq. unit Solution: (b) [latex][\frac{1}{\sqrt{2}}/latex] sq. units Case-Study 2 : Consider two vectors a = [latex]\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}$$ and b = $$\vec{b}=2 \hat{i}-\hat{j}+3 \hat{k}$$.

Based on the above information, answer the following questions:

(i) Projection of $$\vec{a}$$ on $$\vec{b}$$ is
(a) $$\frac{13}{\sqrt{14}}$$
(b) $$\frac{9}{\sqrt{14}}$$
(c) $$\frac{11}{\sqrt{14}}$$
(d) $$\frac{7}{\sqrt{14}}$$
Solution:
(b) $$\frac{9}{\sqrt{14}}$$

(ii) Unit vectors perpendicular to vectors $$\vec{a}$$ and $$\vec{b}$$ are
(a) ± $$\frac{(9 \hat{i}+3 \hat{j}+6 \hat{k})}{\sqrt{126}}$$
(b) ± $$\frac{(9 \hat{i}-4 \hat{j}+5 \hat{k})}{\sqrt{122}}$$
(c) ± $$\frac{(9 \hat{i}+3 \hat{j}-5 \hat{k})}{\sqrt{115}}$$
(d) ± $$\frac{(-7 \hat{i}+3 \hat{j}+5 \hat{k})}{\sqrt{83}}$$
Solution:
(c) ± $$\frac{(9 \hat{i}+3 \hat{j}-5 \hat{k})}{\sqrt{115}}$$

(iii) Area of a parallelogram whose adjacent sides are given by the vectors $$\vec{a}$$ and $$\vec{b}$$ is
(a) $$\sqrt{115}$$ sq. units
(b) $$\sqrt{126}$$ sq. units
(c) $$\sqrt{122}$$ sq. units
(d) $$\sqrt{83}$$ sq. units
Solution:
(a) $$\sqrt{115}$$ sq. units

∴ area of ||gm = $$|\vec{a} \times \vec{b}|$$
= $$\sqrt{115}$$ sq. units

(iv) Area of a parallelogram whose diagonals are $$\vec{a}$$ and $$\vec{b}$$ is
(a) $$\frac{1}{2} \sqrt{83}$$
(b) $$\frac{1}{2} \sqrt{122}$$
(c) $$\frac{1}{2} \sqrt{126}$$
(d) $$\frac{1}{2} \sqrt{115}$$
Solution:
(d) $$\frac{1}{2} \sqrt{115}$$

∴ area of || gm whose diagonals are $$\vec{a}$$ and $$\vec{b}$$
= $$\frac{1}{2}|\vec{a} \times \vec{b}|$$
= $$\frac{\sqrt{115}}{2}$$ sq. units

Chapter 2 Three Dimensional Geometry
Case-Study Based Questions (Solved)

Case-Study 1 :

Consider a straight line joining two points A (- 2, – 1, 3) and B (1, 1, 5).

Based on the above information, answer the following questions:

(i) Direction cosines of line joining the points A and B are
(a) < $$-\frac{3}{\sqrt{17}},-\frac{2}{\sqrt{17}}, \frac{2}{\sqrt{17}}$$ >
(b) < $$\frac{2}{\sqrt{17}}, \frac{3}{\sqrt{17}}, \frac{2}{\sqrt{17}}$$ >
(c) < $$\frac{2}{\sqrt{17}}, \frac{2}{\sqrt{17}}, \frac{3}{\sqrt{17}}$$ >
(d) < $$\frac{3}{\sqrt{17}}, \frac{2}{\sqrt{17}}, \frac{2}{\sqrt{17}}$$ >
Solution:
(d) < $$\frac{3}{\sqrt{17}}, \frac{2}{\sqrt{17}}, \frac{2}{\sqrt{17}}$$ >

Direction ratios of line joining the points A (- 2, – 1, 3) and B (1, 1, 5)
= < 1 – (- 2), 1 – (- 1), 5 – 3 >
i.e. < 3, 2, 2 >
direction cosines of line segment AB are ;
< $$\frac{3}{\sqrt{3^2+2^2+2^2}}, \frac{2}{\sqrt{3^2+2^2+2^2}}, \frac{2}{\sqrt{3^2+2^2+2^2}}$$ >
i.e. < $$\frac{3}{\sqrt{17}}, \frac{2}{\sqrt{17}}, \frac{2}{\sqrt{17}}$$ >

(ii) Cartesian equation of line joining the points A and B is
(a) $$\frac{x-1}{-3}=\frac{y-1}{-2}=\frac{z-5}{2}$$
(b) $$\frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}$$
(c) $$\frac{x+2}{2}=\frac{y+1}{2}=\frac{z-3}{3}$$
(d) $$\frac{x-1}{2}=\frac{y-1}{2}=\frac{z-5}{2}$$
Solution:
(b) $$\frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}$$

The cartesian eqn. of line through the points A and B is given by
$$\frac{x-(-2)}{1-(-2)}=\frac{y-(-1)}{1-(-1)}=\frac{z-3}{5-3}$$
i.e. $$\frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}$$

(iii) Points on the line joining the points A and B which are at a distance of 5 units from the point (1, 3, 3) are
(a) (2, 1, 3), (4, 3 ,- 7)
(b) (- 2, 1, – 3), (4, – 3, – 7)
(c) (-2, – 1, 3), (4, 3, 7)
(d) (2, 1, – 3), (4, – 3, 7)
Solution:
(c) (-2, – 1, 3), (4, 3, 7)

Clearly any point on line joining A and B be given by
$$\frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}$$ = t (say)
i.e. P(3t – 2, 2t – 1, 2t + 3)
Further given |PQ| = 5 where Q (1, 3, 3).
⇒ $$\sqrt{(3 t-2-1)^2+(2 t-1-3)^2+(2 t+3-3)^2}$$ = 5
⇒ (3t – 3)2 + (2t – 4)2 + (2t)2 = 25
⇒ 9(t2 – 2t + 1) + 4(t2 – 4t + 4) + 4t2 = 25
⇒ 17t2 – 34t = 0
⇒ 17t (t – 2)= 0
⇒ t = 0, 2
When t = 0 ;
point P becomes (- 2, – 1, 3)
When t = 2 ;
point P becomes (4, 3, 7)

(iv) The coordinates of the point, where the line joining the points A and B cuts the xz-plane are
(a) (- $$\frac{1}{2}$$, 0, 4)
(b) (0, $$\frac{1}{3}$$, $$\frac{13}{3}$$)
(c) (- $$\frac{13}{2}$$, – 4, 0)
(d) ($$\frac{2}{3}$$, 0, $$\frac{1}{3}$$)
Solution:
(a) (- $$\frac{1}{2}$$, 0, 4)

Let the point Q divides the line segment AB in the ratio k : 1.
Then by section formula,

The coordinates of Q are $$\left(\frac{k-2}{k+1}, \frac{k-1}{k+1}, \frac{5 k+3}{k+1}\right)$$
Now line segment AB meets xz-plane.
∴ y-coordinate of Q is 0
⇒ $$\frac{k-1}{k+1}$$ = 0
⇒ k = 1
Coordinates of Q are (- $$\frac{1}{2}$$, 0, 4).

Case – Study 2:

In the space, a boy is standing at a point (3, 2, 1). He wants to reach a plane 2x – y + z + 1 = 0 in the space.

Based on the above information, answer the following questions:

(i) Equation of path followed by the boy to reach the plane in least possible time is
(a) $$\frac{x+3}{2}=\frac{y-2}{1}=\frac{z-1}{-1}$$
(b) $$\frac{x-3}{2}=\frac{y-2}{-1}=\frac{z-1}{1}$$
(c) $$\frac{x-3}{1}=\frac{y-2}{2}=\frac{z-1}{-1}$$
(d) $$\frac{x-3}{-2}=\frac{y-2}{-1}=\frac{z-1}{-1}$$
Solution:
(b) $$\frac{x-3}{2}=\frac{y-2}{-1}=\frac{z-1}{1}$$

eqn. of plane be
2x – y + z + 1 = 0 ……………………(1)
Now boy reaching the plane in least possible time if he follows the path which is line of shortest distance.

$$\frac{x-3}{2}=\frac{y-2}{-1}=\frac{z-1}{1}$$ …………………..(2)

(ii) Point of intersection of the path followed by the boy for reaching to plane in least time and the plane is
(a) (1, 3, 0)
(b) (- 1, 0, 3)
(c) (3, 0, 1)
(d) (3, – 1, 0)
Solution:
(a) (1, 3, 0)

Let P be the point of intersection of line (2) and plane (1).
Any point on line (2) be (2t + 3, – t + 2, t + 1)
For intersection, P (2t + 3, – t + 2, t + 1)
lies on plane (1)
2 (2t + 3) – (- t + 2) + t + 1 + 1 = 0
⇒ 6t + 6 = 0
⇒ t = – 1
Thus coordinates of point of intersection P are (- 2 + 3, 1 + 2, – 1 + 1) i.e. (1, 3, 0)

(iii) Image of the point (3, 2, 1) in the plane 2x – y + z + 1 = 0 (assuming plane as a mirror) is
(a) (4, 1, 1)
(b) (4, – 1, – 1)
(e) (1, 4, 1)
(d) (- 1, 4, – 1)
Solution:
(d) (- 1, 4, – 1)

Let P be the foot of ⊥ drawn from A (3, 2, 1) to given plane (1).
Then coordinates of P are (1, 3, 0).
Let Q (α, β, γ) be the image of A (3, 2, 1) in plane (1) then P be the mid-point of AQ.

∴ $$\left(\frac{\alpha+3}{2}, \frac{\beta+2}{2}, \frac{\gamma+1}{2}\right)$$ = (1, 3, 0)
i.e. $$\frac{\alpha+3}{2}$$ = 1
⇒ α = – 1;
$$\frac{\beta+2}{2}$$ = 3
⇒ β = 4
and $$\frac{\gamma+1}{2}$$ = 0
⇒ γ = – 1
∴ Coordinates of image of given point are (- 1, 4, – 1).

(iv) Distance of the boy from the plane 2x – y + z + 1 = 0 measured along a straight line path parallel to line $$\frac{x+4}{1}=\frac{y-2}{2}=\frac{z+3}{3}$$ is
(a) 2 $$\sqrt{14}$$ units
(b) 3 $$\sqrt{14}$$ units
(c) 2√7 units
(d) 3√7 units
Solution:
(a) 2 $$\sqrt{14}$$ units

The eqn. of line through a point (3, 2, 1) and || to given line $$\frac{x+4}{1}=\frac{y-2}{2}=\frac{z+3}{3}$$
be $$\frac{x-3}{1}=\frac{y-2}{2}=\frac{z-1}{3}$$ …………………(1)

So any point on line (1) be P (t + 3, 2t + 2, 3t + 1)
Clearly point P lies on plane
2x – y + z + 1 = 0
∴ 2 (t + 3) – (2t + 2) + 3t + 1 + 1 = 0
∴ 3t = – 6
⇒ t = – 2
Thus coordinates of P are (- 2 + 3, – 4 + 2, – 6 + 1)
i.e. (1, – 2, – 5)
∴ |AP| = $$\sqrt{(3-1)^2+(2+2)^2+(1+5)^2}$$
= $$\sqrt{4+16+36}$$
= $$\sqrt{56}$$
= $$2 \sqrt{14}$$ units

Case – Study 3 :

A helicopter ¡s flying in the sky. The coordinates of the helicopter are P (- 2, 3, – 4). A bird is also flying in the sky along the line whose equation is $$\frac{x+2}{2}=\frac{2 y+3}{4}=\frac{3 z+4}{5}$$.

Based on the above information, answer the following questions:

(i) If Q is any point on the given line and PQ is parallel to the plane 4x + 12y – 3z + 1 = 0, then the coordinates of Q are
(a) (4, $$\frac{5}{2}$$, – 2)
(b) (- 4, $$\frac{5}{2}$$ , 2)
(c) (- 4, – $$\frac{5}{2}$$, – 2)
(d) (4, $$\frac{5}{2}$$, 2)
Solution:
(d) (4, $$\frac{5}{2}$$, 2)

Given coordinates of helicopter are P ( – 2, 3, – 4) given eqn. of path of bird be given by
$$\frac{x+2}{3}=\frac{y+\frac{3}{2}}{2}=\frac{z+\frac{4}{3}}{\frac{5}{3}}$$
i.e. $$\frac{x+2}{9}=\frac{y+\frac{3}{2}}{6}=\frac{z+\frac{4}{3}}{5}$$ ………………..(1)
Any point on line (1) be
Q (9t – 2, 6t – $$\frac{3}{2}$$, 5t – $$\frac{4}{3}$$)
direction ratios of line PQ are
< 9t – 2 + 2, 6t – $$\frac{3}{2}$$, 5t – $$\frac{4}{3}$$ ± 4 >
i.e. < 9t, 6t – $$\frac{9}{2}$$, 5t + $$\frac{8}{3}$$ >
Direction ratios of normal to given plane 4x + 12y – 3z + 1 = 0 are < 4, 12, – 3 >
Since PQ is parallel to given plane.
∴ NormaI to plane is ⊥ to line PQ.
Thus, 4 × 9t + 12 × (6t – $$\frac{9}{2}$$) – 3 (5t + $$\frac{8}{3}$$) = 0
36t + 72t – 54 – 15t – 8 = 0
⇒ 93t – 62 = 0
⇒ t = $$\frac{2}{3}$$
∴ Coordinates of Q are $$\left(9 \times \frac{2}{3}-2,6 \times \frac{2}{3}-\frac{3}{2}, 5 \times \frac{2}{3}-\frac{4}{3}\right)$$
i.e. (4, $$\frac{5}{2}$$, 2)

(ii) The distance of the point (- 2, 3, – 4) from the line $$\frac{x+2}{3}=\frac{2 y+3}{4}=\frac{3 z+5}{5}$$ measured parallel to the plane 4x + 12y – 3z + 1 = 0 is
(a) 7.5 units
(b) 8.5 units
(c) 9.5 units
(d) 10.5 units
Solution:
(b) 8.5 units

Clearly any point on given line (1) be Q (9t – 2, 6t – $$\frac{3}{2}$$, 5t – $$\frac{4}{3}$$)
∴ direction ratios of line PQ are < 9t – 2 + 2, 6t – $$\frac{3}{2}$$ – 3, 5t – $$\frac{4}{3}$$ + 4 >
i.e. < 9t, 6t – $$\frac{9}{2}$$, 5t + $$\frac{8}{3}$$ >
Now direction ratios of normal to given plane are < 4, 12, – 3 >
Since PQ is parallel to given plane.
∴ Normal to plane is ⊥ to line PQ.
∴ 9t × 4 + (6t – $$\frac{9}{2}$$) 12 – 3 (5t + $$\frac{8}{3}$$) = 0
⇒ 36t + 72t – 54 – 15t – 8 = 0
⇒ 93t = 62
⇒ t = $$\frac{2}{3}$$
∴ Coordinates of Q are (4, $$\frac{5}{2}$$, 2)
Thus |PQ| = $$\sqrt{(4+2)^2+\left(\frac{5}{2}-3\right)^2+(2+4)^2}$$
= $$\sqrt{36+\frac{1}{4}+36}$$
= $$\sqrt{\frac{289}{4}}$$
= $$\frac{17}{2}$$ sq. units

(iii) The equation of a plane passing through the point (- 2, 3, – 4) and perpendicular to the line $$\frac{x+2}{3}=\frac{2 y+3}{4}=\frac{3 z+5}{5}$$ is
(a) 3 (x + 2) + 2 (y – 3) + $$\frac{5}{3}$$ (z + 4) = 0
(b) 3 (x + 2) + 4 (y – 3) + 5 (z + 4) = 0
(c) – 3 (x + 2) + 4 (y – 3) + 5 (z + 4) = 0
(d) none of these
Solution:
(a) 3 (x + 2) + 2 (y – 3) + $$\frac{5}{3}$$ (z + 4) = 0

Direction ratios of line (1) are < 9, 6, 5 >
Thus eqn. of plane passing through the point (- 2, 3, – 4) and ⊥ to line (1) be given by
9 (x + 2) + 6 (y – 3) + 5 (z + 4) = 0
3 (x + 2) + 2(y + 3) + $$\frac{5}{3}$$ (z + 4) = 0

(iv) The direction cosines of the plane 4x + 12y – 3z + 1 = 0 are
(a) < $$\frac{4}{13}, \frac{12}{13}, \frac{-3}{13}$$ >
(b) < $$\frac{-4}{13}, \frac{-12}{13}, \frac{-3}{13}$$ >
(c) < $$\frac{4}{13}, \frac{12}{13}, \frac{3}{13}$$ >
(d) none of these
Solution:
(a) < $$\frac{4}{13}, \frac{12}{13}, \frac{-3}{13}$$ >

Direction ratios of normal to given plane are < 4, 12, – 3 >
∴ direction cosine of normal to plane are :
< $$\frac{4}{\sqrt{16+144+9}}, \frac{12}{\sqrt{16+144+9}}, \frac{-3}{\sqrt{16+144+9}}$$ >
i.e. < $$\frac{4}{13}, \frac{12}{13}, \frac{-3}{13}$$ >

Case – Study 4 :

Observe the diagram shows alongside in which two lines l1 and l2 are shown in space and $$\overline{\mathbf{A B}}$$ is the only line which is perpendicular to both the lines l1</sub and l2</sub.

Based on the above information, answer the following questions:

(i) Which of the following is true?
(a) l1 and l2 are parallel lines
(b) l1 and l2 are intersecting lines
(c) l1 and l2 are skew lines
(d) none of these
Solution:
(c) l1 and l2 are skew lines

Since l1 and l2 are neither intersecting nor parallel so l1 and l2 are skew lines.
[as AB ¡s the only line ⊥ to both l1 and l2]

(ii) If equations of l1 and l2 are respectively $$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$$ and $$\frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$$ then a vector perpendicular to both the lines is given by
(a) $$\hat{i}+2 \hat{j}-\hat{k}$$
(b) $$-\hat{i}+2 \hat{j}-\hat{k}$$
(c) $$\hat{i}+\mathbf{2} \hat{j}+\hat{k}$$
(d) $$2 \hat{i}+\hat{j}+\hat{k}$$
Solution:

Given eqn. of lines are ;
$$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$$ ………………(1)
$$\frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$$ …………….(2)
Clearly direction ratios of line (1) are < 2, 3, 4 >
D’ratios of line (2) are < 3, 4, 5 >
For option (b) ;
the vector $$-\hat{i}+2 \hat{j}-\hat{k}$$ is || to line having direction ratios < – 1, 2, – 1 >
Here 2 × (- 1) + 3 × 2 + 4 × (- 1) = 0
and 3 × (- 1) + 4 × 2 + 5 × (- 1) = 0

(iii) Two given points on the lines l1 and l2 are
(a) (1, 2, 3) and (2, 4, 5)
(b) (2, 3, 4) and (3, 4, 5)
(c) (- 1, – 2, – 3) and (- 2, – 4, – 5)
(d) none of these
Solution:
(a) (1, 2, 3) and (2, 4, 5)

Clearly the lines (1) and (2) passing through the point A (1, 2,3) and B (2, 4, 5).

(iv) What is the magnitude of $$\overrightarrow{\mathbf{A B}}$$ i.e. $$|\overrightarrow{\mathbf{A B}}|$$ ?
(a) √6 units
(b) $$\frac{1}{\sqrt{6}}$$ units
(e) 2√6 units
(d)  units
Solution:
(b) $$\frac{1}{\sqrt{6}}$$ units

The vector eqns. of given lines (1) and (2)

Chapter 3 Applications of Integrals
Case – Study Based Questions (Solved)

Case – Study 1 :

Consider the following equations of curves x2 = y and y = x.

Based on the above information, answer the following questions :

(i) The points of intersection of both the curves are
(a) (0, 0), (2, 2)
(b) (0, 0), (- 2, 2)
(c) (0, 0), (1, 1)
(d) (0,0, (- 1, 1)
Solution:
(c) (0, 0), (1, 1)

eqns. of given curves are ;
x2 = y ……………………(1)
and y = x ……………….(2)
both curves (1) and (2) intersects when
x2 = x
⇒ x (x – 1) = 0
⇒ x = 0, 1
⇒ y = 0, 1
Thus points of intersection of both curves are (0, 0) and (1, 1).

(ii) The graph of the given curves is
(a)

(b)

(c)

(d)
Solution:
(d)

eqn. (1) represents an upward parabola with vertex (0, 0)
and y = x be a line passing through (0, 0) making a slope of 45° with +ve direction of x-axis.

(iii) The value of the integral $$\int_0^1$$ x2 dx is
(a) 1
(b) $$\frac{1}{3}$$
(c) $$\frac{2}{3}$$
(d) $$\frac{1}{4}$$
Solution:
(b) $$\frac{1}{3}$$

$$\left.\int_0^1 x^2 d x=\frac{x^3}{3}\right]_0^1=\frac{1}{3}$$

(iv) The area bounded by the curves x2 = y and y = x is
(a) $$\frac{1}{2}$$ sq. unit
(b) $$\frac{1}{3}$$ sq. unit
(c) $$\frac{1}{4}$$ sq. unit
(d) $$\frac{1}{6}$$ sq. unit
Solution:
(d) $$\frac{1}{6}$$ sq. unit

∴ Required area = $$\int_0^1$$ [x – x2]
= $$\left.\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1$$
= $$\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$$ sq. units.

Case – Study 2:

Three friends A, B and C are marking the boundary of a triangular field by putting 3 flags at the corner points of the field as shown in figure given below:

Based on the above information, answer the following questions :

(i) The correct equation representing the boundary BA is
(a) y = x + 5
(b) y = $$\frac{5}{2}$$ x – 5
(c) y = 5 – $$\frac{5}{2}$$ x
(d) y = – 5 – $$\frac{5}{2}$$ x
Solution:
(b) y = $$\frac{5}{2}$$ x – 5

The eqn. representing the boundary BA is given by
y – 0 = $$\frac{5-0}{4-2}$$ (x – 2)
2y = 5x – 10
y = $$\frac{5}{2}$$ x – 5

(ii) The correct equation representing the boundary AC is
(a) y = – x – 9
(b) y = 9 + x
(c) y = 9 – x
(d) y = x – 9
Solution:
(c) y = 9 – x

The eqn. representing the boundary AC is given by
y – 5 = $$\frac{3-5}{6-4}$$ (x – 4)
⇒ y – 5 = – (x – 4)
⇒ y = – x + 4 + 5 = 9 – x

(iii) The correct equation representing the boundary BC is
(a) y = $$\frac{x}{2}-\frac{3}{4}$$
(b) y = $$\frac{3}{2} x-\frac{3}{2}$$
(c) y = $$\frac{3}{4} x-\frac{3}{2}$$
(d) y = $$\frac{3}{2}-\frac{3}{4} x$$
Solution:
(c) y = $$\frac{3}{4} x-\frac{3}{2}$$

The eqn. representing the boundary BC is given by
y – 0 = $$\frac{3-0}{6-2}$$ (x – 2)
⇒ y = $$\frac{3}{2}$$ (x – 2)
= $$\frac{3 x}{4}-\frac{3}{2}$$

(iv) Area of triangular field is given by
(a) 7 sq. units
(b) 6 sq. units
(c) 8 sq. units
(d) 9 sq. units
Solution:

∴ area of triangular field = area of region BAEB + area of region EACE