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(t^{2} – 2t + 1) + 4(t^{2} – 4t + 4) + 4t^{2} = 25

⇒ 17t^{2} – 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 l_{1} and l_{2} are shown in space and \(\overline{\mathbf{A B}}\) is the only line which is perpendicular to both the lines l_{1</sub and l2</sub.}

Based on the above information, answer the following questions:

(i) Which of the following is true?

(a) l_{1} and l_{2} are parallel lines

(b) l_{1} and l_{2} are intersecting lines

(c) l_{1} and l_{2} are skew lines

(d) none of these

Solution:

(c) l_{1} and l_{2} are skew lines

Since l_{1} and l_{2} are neither intersecting nor parallel so l_{1} and l_{2} are skew lines.

[as AB ¡s the only line ⊥ to both l_{1} and l_{2}]

(ii) If equations of l_{1} and l_{2} 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 l_{1} and l_{2} 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 x^{2} = 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 ;

x^{2} = y ……………………(1)

and y = x ……………….(2)

both curves (1) and (2) intersects when

x^{2} = 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 \(\int_0^1\) x^{2} dx is

(a) 1

(b) \(\frac{1}{3}\)

(c) \(\frac{2}{3}\)

(d) \(\frac{1}{4}\)

Solution:

(b) \(\frac{1}{3}\)

(iv) The area bounded by the curves x^{2} = 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 – x^{2}]

= \(\left.\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1\)

= \(\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\) 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