Chapter 14 - Equation of a Line Exercise Ex. 14(B)

Question 1

Find the slope of the line whose inclination is:

(i) 0 (ii) 30

(iii) 72 30' (iv) 46

Solution 1

(i) Slope = tan 0^o = 0

(ii) Slope = tan 30^o = 

(iii) Slope = tan 72^o 30' = 3.1716

(iv) Slope = tan 46^o = 1.0355

Question 2

Find the inclination of the line whose slope is:

(i) 0 (ii) 

(iii) 0.7646 (iv) 1.0875

Solution 2

(i) Slope = tan  = 0

 = 0^o

(ii) Slope = tan  = 

 = 60^o

(iii) Slope = tan  = 0.7646

 = 37^o 24'

(iv) Slope = tan  = 1.0875

 = 47^o 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 3

We know:

Slope = 

(i) Slope = 

(ii) Slope = 

(iii) Slope = 

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 4

(i) Slope of AB = 

Slope of the line parallel to AB = Slope of AB = 1

(ii) Slope of AB = 

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 5

(i) Slope of AB = 

Slope of the line perpendicular to AB =  

(ii) Slope of AB = 

Slope of the line perpendicular to AB =  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 6

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

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

Since, the lines are parallel.

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 7

Slope of the line passing through (-4, -2) and (2, -3) = 

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

Since, the lines are perpendicular.

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 8

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

 

It can be seen that:

Hence, AB  AC.

Thus, 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 9

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

Since, slope of AB = slope of CD

 

Therefore AB || CD

 

 

 

Since, slope of BC = slope of DA

Therefore, BC || DA

 

Hence, 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 10

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

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

 

Co-ordinates of P are

 

Co-ordinates of Q are

 

Co-ordinates of R are

 

Co-ordinates of S are

 

 

Since, slope of PQ = Slope of RS, PQ || RS.

 

 

 

Since, slope of QR = Slope of SP, QR || SP.

Hence, PQRS is a parallelogram.

Question 11

Show that the points P (a, b + c), Q (b, c + a) and R (c, a + b) are collinear.

Solution 11

The points P, Q, R will be collinear if slope of PQ and QR is the same.

 

 

 

Hence, the points P, Q, and R are collinear.

Question 12

Find x, if the slope of the line joining (x, 2) and (8, -11) is.

Solution 12

Let A = (x, 2) and B = (8, -11)

Slope of AB = 

Question 13

The side AB of an equilateral triangle ABC is parallel to the x-axis. Find the slope of all its sides.

Solution 13

We know that the slope of any line parallel to x-axis is 0.

Therefore, slope of AB = 0

 

Since, ABC is an equilateral triangle, 

Slope of AC = tan 60^o =

 

Slope of BC = -tan 60^o = -

Question 14

The side AB of a square ABCD is parallel to the x-axis. Find the slopes of all its sides.

Also, find:

(i) the slope of the diagonal AC,

(ii) the slope of the diagonal BD.

 

Solution 14

We know that the slope of any line parallel to x-axis is 0.

Therefore, slope of AB = 0

 

As CD || AB, slope of CD = Slope of AB = 0

 

As BC  AB, slope of BC = 

As AD  AB, slope of AD = 

 

(i) The diagonal AC makes an angle of 45^o with the positive direction of x axis.

(ii) The diagonal BD makes an angle of -45^o with the positive direction of x axis.

Question 15

A (5, 4), B (-3, -2) and C (1, -8) are the vertices of a triangle ABC. Find:

(i) the slope of the altitude of AB,

(ii) the slope of the median AD, and

(iii) the slope of the line parallel to AC.

Solution 15

Given, A (5, 4), B (-3, -2) and C (1, -8) are the vertices of a triangle ABC.

 

(i) Slope of AB = 

Slope of the altitude of AB = 

 

(ii) Since, D is the mid-point of BC.

Co-ordinates of point D are

Slope of AD = 

 

(iii) Slope of AC = 

Slope of line parallel to AC = Slope of AC = 3

Question 16

The slope of the side BC of a rectangle ABCD is. Find:

(i) the slope of the side AB,

(ii) the slope of the side AD.

Solution 16

(i) Since, BC is perpendicular to AB,

Slope of AB = 

 

(ii) Since, AD is parallel to BC,

Slope of AD = Slope of BC = 

Question 17

Find the slope and the inclination of the line AB if:

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

(ii) A = (0, ) and B = (3, 0)

(iii) A = (-1, 2) and B = (-2, )

Solution 17

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

Slope of AB = 

Inclination of line AB =  = 45^o

 

(ii) A = (0, ) and B = (3, 0)

Slope of AB = 

Inclination of line AB =  = 30^o

 

(iii) A = (-1, 2) and B = (-2, )

Slope of AB = 

Inclination of line AB =  = 60^o

Question 18

The points (-3, 2), (2, -1) and (a, 4) are collinear. Find a.

Solution 18

Given, points A (-3, 2), B (2, -1) and C (a, 4) are collinear.

 Slope of AB = Slope of BC

Question 19

The points (K, 3), (2, -4) and (-K + 1, -2) are collinear. Find K.

Solution 19

Given, points A (K, 3), B (2, -4) and C (-K + 1, -2) are collinear.

 Slope of AB = Slope of BC

Question 20

Plot the points A (1, 1), B (4, 7) and C (4, 10) on a graph paper. Connect A and B, and also A and C.

Which segment appears to have the steeper slope, AB or AC?

Justify your conclusion by calculating the slopes of AB and AC.

Solution 20

 

From the graph, clearly, AC has steeper slope.

 

Slope of AB = 

Slope of AC = 

The line with greater slope is steeper. Hence, AC has steeper slope.

Question 21

Find the value(s) of k so that PQ will be parallel to RS. Given:

(i) P (2, 4), Q (3, 6), R (8, 1) and S (10, k)

(ii) P (3, -1), Q (7, 11), R (-1, -1) and S (1, k)

(iii) P (5, -1), Q (6, 11), R (6, -4k) and S (7, k²)

Solution 21

Since, PQ || RS,

Slope of PQ = Slope of RS

 

(i) Slope of PQ = 

Slope of RS = 

 

(ii) Slope of PQ = 

Slope of RS = 

 

(iii) Slope of PQ = 

Slope of RS =