Selina Solutions Concise Maths Class 10 Chapter 14 Equation of a Line

This Chapter is one of the important chapters of Class 10 ICSE. The topics covered under this Chapter are basic concept of a straight line equation, inclination of a line, concept of slope (gradient), slope of a straight line passing through two given fixed points, slopes of parallel and straight lines, various forms of the equations of straight lines and finding x/y -intercepts of a given line. Also, creating a good understanding of these concepts lays a strong foundation in higher classes. Students wishing to get a good grip over the chapters of Selina Solutions for Class 10 Mathematics can download the PDF of solutions prepared by subject experts at MPS. Students can now access the Selina Solutions for Class 10 Mathematics Chapter 14 Equation of a Line from the links enclosed below.

Selina Solutions Concise Maths Class 10 Chapter 14 Equation of a Line

Exercise 14(A) Solutions

Exercise 14(B) Solutions

Exercise 14(C) Solutions

Exercise 14(D) Solutions

Exercise 14(E) Solutions

Selina Solutions Concise Maths Class 10 Chapter 14 Equation of a Line Exercise 14(B)

Exercise 14(B) Page No: 190

Question 1. Find the slope of the line whose inclination is:

(i) 0o (ii) 30o

(iii) 72o 30′ (iv) 46o

Solution:

We know that, the slope of a line is given by the tan of its inclination.

(i) Slope = tan 0o = 0

(ii) Slope = tan 30o = 1/ √3

(iii) Slope = tan 72o 30′ = 3.1716

(iv) Slope = tan 46o = 1.0355

Question 2. Find the inclination of the line whose slope is:

(i) 0 (ii) √3

(iii) 0.7646 (iv) 1.0875

Solution:

(i) Slope = tan θ = 0

⇒ θ = 0o

(ii) Slope = tan θ = √3

⇒ θ = 60o

(iii) Slope = tan θ = 0.7646

⇒ θ = 37o 24′

(iv) Slope = tan θ = 1.0875

⇒ θ = 47o 24′

Question 3. Find the slope of the line passing through the following pairs of points:

(i) (-2, -3) and (1, 2)

(ii) (-4, 0) and origin

(iii) (a, -b) and (b, -a)

Solution:

We know that,

Slope = (y2 – y1)/ (x2 – x1)

(i) Slope = (2 + 3)/ (1 + 2) = 5/3

(ii) Slope = (0 – 0)/ (0 + 4) = 0

(iii) Slope = (-a + b)/ (b – a) = 1

Question 4. Find the slope of the line parallel to AB if:

(i) A = (-2, 4) and B = (0, 6)

(ii) A = (0, -3) and B = (-2, 5)

Solution:

(i) Slope of AB = (6 – 4)/ (0 + 2) = 2/2 = 1

Hence, slope of the line parallel to AB = Slope of AB = 1

(ii) Slope of AB = (5 + 3)/ (-2 – 0) = 8/-2 = -4

Hence, slope of the line parallel to AB = Slope of AB = -4

Question 5. Find the slope of the line perpendicular to AB if:

(i) A = (0, -5) and B = (-2, 4)

(ii) A = (3, -2) and B = (-1, 2)

Solution:

(i) Slope of AB = (4 + 5)/ (-2 – 0) = -9/2

Slope of the line perpendicular to AB = -1/Slope of AB = -1/(-9/2 ) = 2/9

(ii) Slope of AB = (2 + 2)/ (-1 – 3) = 4/-4 = -1

Slope of the line perpendicular to AB = -1/slope of AB = -1/-1 = 1

Question 6. The line passing through (0, 2) and (-3, -1) is parallel to the line passing through (-1, 5) and (4, a). Find a.

Solution:

Slope of the line passing through (0, 2) and (-3, -1) = (-1 – 2)/ (-3 – 0) = -3/-3 = 1

Slope of the line passing through (-1, 5) and (4, a) = (a – 5)/ (4 + 1) = (a -5)/ 5

As, the lines are parallel.

The slopes must be equal.

1 = (a – 5)/ 5

a – 5 = 5

a = 10

Question 7. The line passing through (-4, -2) and (2, -3) is perpendicular to the line passing through (a, 5) and (2, -1). Find a.

Solution:

Slope of the line passing through (-4, -2) and (2, -3) = (-3 + 2)/ (2 + 4) = -1/6

Slope of the line passing through (a, 5) and (2, -1) = (-1 – 5)/ (2 – a) = -6/ (2 –a)

As, the lines are perpendicular we have

-1/6 = -1/(-6/ 2 – a)

-1/6 = (2 – a)/6

2 – a = -1

a = 3

Question 8. Without using the distance formula, show that the points A (4, -2), B (-4, 4) and C (10, 6) are the vertices of a right-angled triangle.

Solution:

Given points are A (4, -2), B (-4, 4) and C (10, 6).

Calculating the slopes, we have

Slope of AB = (4 + 2)/ (-4 – 4) = 6/ -8 = -3/4

Slope of BC = (6 – 4)/ (10 + 4) = 2/ 14 = 1/7

Slope of AC = (6 + 2)/ (10 – 4) = 8/ 6 = 4/3

It’s clearly seen that,

Slope of AB = -1/ Slope of AC

Thus, AB ⊥ AC.

Therefore, the given points are the vertices of a right-angled triangle.

Question 9. Without using the distance formula, show that the points A (4, 5), B (1, 2), C (4, 3) and D (7, 6) are the vertices of a parallelogram.

Solution:

Given points are A (4, 5), B (1, 2), C (4, 3) and D (7, 6).

Slope of AB = (2 – 5)/ (1 – 4) = -3/ -3 = 1

Slope of CD = (6 – 3)/ (7 – 4) = 3/3 = 1

As the slope of AB = slope of CD

So, we can conclude AB || CD

Now,

Slope of BC = (3 – 2)/ (4 – 1) = 1/3

Slope of DA = (6 – 5)/ (7 – 4) = 1/3

As, slope of BC = slope of DA

Hence, we can say BC || DA

Therefore, ABCD is a parallelogram.

Question 10. (-2, 4), (4, 8), (10, 7) and (11, -5) are the vertices of a quadrilateral. Show that the quadrilateral, obtained on joining the mid-points of its sides, is a parallelogram.

Solution:

Let the given points be A (-2, 4), B (4, 8), C (10, 7) and D (11, -5).

And, let P, Q, R and S be the mid-points of AB, BC, CD and DA respectively.

So,

Selina Solutions Concise Class 10 Maths Chapter 14 ex. 14(B) - 1