# Class 11 RD Sharma Solutions – Chapter 23 The Straight Lines- Exercise 23.1 | Set 1

### Question 1. Find the slope of the lines which make the following angles with the positive direction of x-axis:

### (i) -π/4

**Solution:**

We have

θ = -π/4

Let the slope of the line be ‘m’

We know slope m = tan θ

m = tan (-π/4)

= -1

Therefore, Slope = -1

### (ii) 2π/3

**Solution:**

We have

θ = 2π/3

Let the slope of the line be ‘m’

We know slope m = tan θ

m = tan (2π/3)

= tan (π – π/3)

= -tan π/3

= √3

Therefore, Slope = √3

### (iii) 3π/4

**Solution:**

We have

θ = 3π/4

Let the slope of the line be ‘m’

We know slope m = tan θ

m = tan (3π/4)

= tan (π – π/4)

= -tan π/4

= -1

Therefore, Slope = -1

### (iv) π/3

**Solution:**

We have

θ = π/3

Let the slope of the line be ‘m’

We know slope m = tan θ

m = tan (π/3)

= tan π/3

= √3

Therefore, Slope = √3

### Question 2. Find the slope of a line passing through the following points:

### (i) (-3, 2) and (1, 4)

**Solution:**

Let ‘m’ be the slope of the line

So, the slope of line m = [4 – 2] / [1 – (-3)] [Using: m = [y

_{2 }– y_{1}] / [x_{2 }– x_{1}]= 2/4

= 1/2

### (ii) (at^{2}_{1}, 2at_{1}) and (at^{2}_{2}, 2at_{2})

**Solution:**

Slope of line ‘m’ = [2at

_{2 }– 2at_{1}] / [at^{2}_{2 }– at^{2}_{1}]= [2at

_{2 }– t_{1}] / [at^{2}_{2 }– t^{2}_{1}]= 2a(t

_{2 }– t_{1})/a(t_{2 }– t_{1})(t_{2 }+ t_{1})= 2/(t

_{2 }+ t_{1})

### (iii) (3, -5) and (1, 2)

**Solution:**

Slope of line m = [2 – (-5)] / [1 – 3] [Using: m = [y

_{2 }– y_{1}] / [x_{2 }– x_{1}]= 7/-2

= -7/2

### Question 3. State whether the two lines in each of the following are parallel, perpendicular, or neither.

### (i) Through (5, 6) and (2, 3); through (9, -2) and (6, -5)

**Solution:**

Let A(5, 6), B(2, 3), C(9, -2) and D(6, -5) be the given points. Then

m1 = Slope of the line AB

= [3 – 6] / [2 – 5]

= -3/-3

= 1

m2 = Slope of the line CD

= [-5 – (-2)] / [6 – 9]

= -3/3

= -1

Clearly, m1 m2 = -1

This shows that AB is

perpendicularto CD

### (ii) Through (9, 5) and (-1, 1); through (3, -5) and (8, -3)

**Solution:**

Let A(9, 5), B(-1, 1), C(3, -5) and D(8, -3) be the given points. Then

m1 = Slope of the line AB

= [1 – 5] / [-1 – 9]

= -4/-10

= 2/5

m2 = Slope of the line CD

= [-3 – (-5)] / [8 – 3]

= 2/5

Clearly, m1 = m2

This shows that AB is parallel to CD

### (iii) Through (6, 3) and (1, 1); through (-2, 5) and (2, -5)

**Solution:**

Let A(6, 3), B(1, 1), C(-2, 5) and D(2, -5) be the given points. Then

m1 = Slope of the line AB

= [1 – 3] / [1 – 6]

= -2/-5 = 2/5

m2 = Slope of the line CD

= [-5 – 5] / [2 – (-2)]

= 10/4 = -5/2

Clearly, m1 m2 = -1

This shows that AB is perpendicular to CD

### (iv) Through (3, 15) and (16, 6); through (-5, 3) and (8, 2)

**Solution:**

Let A(3, 15), B(16, 6), C(-5, 3) and D(8, 2) be the given points. Then

m1 = Slope of the line AB

= [6 – 15] / [16 – 3]

= -9/13

m2 = Slope of the line CD

= [2 – 3] / [8 – (-5)]

= -1/13

Clearly we don’t see any relation between m1 and m2

So, neither parallel nor perpendicular

### Question 4. Find the slopes of a line

### (i) Which bisects the first quadrant angle

**Solution:**

We have Line bisects the first quadrant

We know if the line bisects in the first quadrant, then

the angle must be between line and the positive direction of x-axis.

Since, angle = 90°/2 = 45°

Using the formula,

Slope of the line, m = tan θ

Slope of the line for a given angle is m = tan 45°

So, m = 1

∴ The slope of the line is 1.

### (ii) Which makes an angle of 30° with the positive direction of y-axis measured anticlockwise.

**Solution:**

We have, The line make an angle 30° with the positive direction of y-axis.

We know, angle between line and positive side of axis is 90° + 30° = 120°

Using formula,

Slope of the line, m = tan θ

Slope of the line for the given angle is m = 120°

So, m =- √3

∴ The slope of the line is – √3.

### Question 5. Using the method of slope, show that the following points are collinear

### (i) A (4, 8), B (5, 12), C (9, 28)

**Solution:**

Using the formula,

Slope of the line = [y

_{2}– y_{1}] / [x_{2}– x_{1}]Slope of the line AB = [12 – 8] / [5 – 4]

= 4/1 = 4

Slope of the line BC = [28 – 12] / [9 – 5]

= 16/4 = 4

Slope of the line CA = [8 – 28] / [4 – 9]

= -20/-5 = 4

Clearly, AB = BC = CA

∴ The given points are collinear.

### (ii) A (16, -18), B (3, -6), C (-10, 6)

**Solution:**

Using the formula,

Slope of the line = [y

_{2}– y_{1}] / [x_{2}– x_{1}]Slope of the line AB = [-6 – (-18)] / [3 – 16]

= 12/-13 = -12/13

Slope of the line BC = [6 – (-6)] / [-10 – 3]

= 12/-13 = -12/13

Slope of the line CA = [6 – (-18)] / [-10 – 16]

= 12/-13 = -12/13

Clearly, AB = BC = CA

∴ The given points are collinear.

### Question 6. What is the value of y so that the line through (3, y) and (2, 7) is parallel to the line through (-1, 4) and (0, 6)?

**Solution:**

Using the formula,

Slope of the line = [y

_{2}– y_{1}] / [x_{2}– x_{1}]Slope of the line joining point (-1, 4) and (0, 6) is

m

_{1 }=_{ }[6 – 4] / [0 – (-1)]= 2

Slope of the line joining point (3, y) and (2, 7) is

m

_{2}= [7 – y] / [2 – 3]= y – 7

As the two lines are parallel m

_{1}= m_{2}⇒ 2 = y – 7

⇒ y = 9

### Question 7. What can be said regarding a line if its slope is

### (i) zero

**Solution:**

If slope = tanθ = 0

⇒ tanθ = tan0

⇒ θ = 0

When the slope of a line is zero then the line is parallel to x-axis.

### (ii) Positive

**Solution:**

When the slope is positive then tanθ is positive and θ is acute angle

thus the line makes an acute angle (0<θ<π/2) with the positive x-axis.

### (iii) Negative

**Solution:**

When the slope is negative then tanθ is negative and θ is obtuse angle

thus the line makes an obtuse angle (θ>π/2) with the positive x-axis.

### Question 8. Show that the line joining (2, -3) and (-5, 1) is parallel to the line joining (7, -1) and (0, 3).

**Solution:**

Using the formula,

Slope of the line = [y

_{2}– y_{1}] / [x_{2}– x_{1}]Slope of the line joining point (2, -3) and (-5, 1) is

m

_{1}= [1 – (-3)] / [(-5) – 2]= 4/(-7) = -(4/7)

Slope of the line joining point (7, -1) and (0, 3) is

m

_{2}= [3 – (-1)] / [ 0 – 7]= 4/(-7) = -(4/7)

Since m

_{1 }= m_{2}, the two lines are parallel

### Question 9. Show that the line joining (2, -5) and (-2, 5) is perpendicular to the line joining (6, 3) and (1, 1).

**Solution:**

Using the formula,

Slope of the line = [y

_{2}– y_{1}] / [x_{2}– x_{1}]Slope of the line joining point (2, -5) and (-2, 5) is

m

_{1}= [5 – (-5)] / [-2 – 2]= -5/2

Slope of the line joining point (6, 3) and (1, 1) is

m

_{2}= [1 – 3] / [1 – 6]= 2/5

m

_{1 }× m_{2 }= [-5/2] × [2/5]= -1

∴ The two given lines are perpendicular to each other.

### Question 10. Without using Pythagoras theorem, show that the points A (0, 4), B (1, 2), and C (3, 3) are the vertices of a right-angled triangle.

** Solution:**

Using the formula,

Slope of the line = [y

_{2}– y_{1}] / [x_{2}– x_{1}]Slope of the line AB = [2 – 4] / [1 – 0]

= -2

Slope of the line BC = [3 – 2] / [3 – 1]

= 1/2

Slope of AB slope of BC = (-2) × (1/2) = -1

∴ Angle between AB and BC = π/2

∴ ABC are the vertices of a right-angled triangle.

### Question 11. Prove that the points (-4, -1), (-2, -4), (4, 0), and (2, 3) are the vertices of a rectangle.

**Solution:**

Let A(-4, -1), B(-2, -4), C(4, 0) and D(2, 3) be the vertices of the rectangle

Slope of the line AB = [-4 + 1] / [-2 + 4]

m1 = -3/2

Slope of the line BC = [0 + 4] / [4 + 2]

m2 = 2/3

Slope of the line CD = [3 – 0] / [2 – 4]

m3 = -3/2

Slope of the line AD = [3 + 1] / [2 + 4]

m4 = 2/3

We see m1 = m3 & m2 = m4

∴ AB || CD & BC || AD

Also, m1 × m2 = -1, m2 × m3 = -1 & m3 × m4 = -1

∴AB⊥BC, BC⊥CD and CD⊥AD

Thus given set of points are the vertices of a rectangle.