Class 11 RD Sharma Solutions – Chapter 23 The Straight Lines- Exercise 23.1 | Set 2

Question 12. If three points A(h, 0), P(a, b) and B(0, k) lie on a line, show that a/h + b/k = 1

Solution:

If the given points lie on a line then we can say that these points have same slope

∴ Slope of AP = Slope of PB = Slope of AB

⇒ [b – 0] / [a – h] = [k – b] / [0 – a] = [k – 0] / [0 – h] 

⇒ [b – 0] / [a – h] = [k – b] / [0 – a] 

⇒ -ab = (k – b)(a – h)

⇒ -ab = ka- kh – ab + bh

⇒ ka + bh = kh         

Dividing both side by kh we get

⇒ a/h + b/k = 1 

Hence Proved

Question 13. The slope of a line is double of the slope of another line. If tangents of the angle between them is 1/3, find the slope of other line. 

Solution:

Let the slope of given lines be m1 and m2

According to question m1 = m and m2 = 2m 

 tan θ = (m1 – m2)/(1 + m1m2)

Case I:

⇒ 1/3 = (m – 2m)/(1 + 2m²)

⇒ 1/3 = (-m)/(1 + 2m²)

⇒ 1 + 2m² = -3m

⇒ 2m² + 3m + 1 = 0

⇒ 2m² + 2m + m + 1 = 0

⇒ 2m(m + 1) + (m + 1) = 0

⇒ (2m + 1)(m + 1) = 0

m = -1, -1/2

Case II:

⇒ 1/3 = (2m – m)/(1 + 2m²)

⇒ 1/3 = (m)/(1 + 2m²)

⇒ 1 + 2m² = 3m

⇒ 2m² – 3m + 1 = 0

⇒ 2m² – 2m – m + 1 = 0

⇒ 2m(m – 1) – (m – 1) = 0

⇒ (2m – 1)(m – 1) = 0

m = 1, 1/2

Question 14. Consider the following population and year graph:

Find the slope of the line AB and using it, find what will be the population in the year 2010.

Solution:

Using the formula,

Slope of the line = [y₂ – y₁] / [x₂ – x₁]

Slope of the line AB = [97 – 92] / [1995 – 1985]

= 5/10 = 1/2

So, the population (P) in 2010 can be find using the slope of AC

Slope of the line AC = [P – 92] / [2010 – 1985]

= (P – 92) / 25

According to question :

 ⇒ (P – 92) / 25 = 1/2 = Slope of AB

 ⇒(P – 92) =25/2

 ⇒ 2P – 184 = 25

 ∴ P = 209/2 = 104.50

Question 15. Without using the distance formula, show that points (-2,-1), (4,0), (3,3), (-3,2) are the vertices of a parallelogram.

Solution:

Let P (-2,-1), Q (4,0), R (3,3) and S (-3,2) be the vertices of a quadrilateral

Using the formula,

Slope of the line = [y₂ – y₁] / [x₂ – x₁]

Slope of the line PQ = [0 – (-1)] / [4 – (-2)]

= 1/6

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

= -3

Slope of the line RS = [2 – 3] / [-3 – 3]

= 1/6

Slope of the line RP = [2 – (-1)] / [-3 – (-2)]

= -3

We see that the slope of opposite side of the quadrilateral PQRS are equal.

Hence, the quadrilateral PQRS is a parallelogram.

Question 16. Find the angle between the x-axis and the line joining the points (3,-1) and (4,-2).

Solution:

Slope of the line segment joining the points (3,-1) and (4,-2) is 

m₁ = [y₂ – y₁] / [x₂ – x₁]

= [-2 – (-1)] / [4 – 3]

= -1

Slope of x-axis is 0

 m₂ = 0

If θ is the angle between x-axis and the line segment then

tanθ = [m₁ – m₂] / [1 + m₁m₂]

= [-1 – 0] / [1 + (-1)(0)]

= -1

∴ θ = 135°

Question 17. Line through the points (-2,6) and (4,8) is perpendicular to the line through the points (8,12) and (x,24). Find the value of x.

Solution:

We have,

Slope of the line = [y₂ – y₁] / [x₂ – x₁]

The slope of the line joining the points (-2,6) and (4,8) is 

m₁ = [8 – 6] / [4 – (-2)]

= 2/6 = 1/3

The slope of the line joining the points (8,12) and (x,24) is 

m₂ = [24 – 12] / [x – 8]

= 12/(x – 8)

The lines are perpendicular to each other

m₁ × m₂ = -1

⇒ (1/3) × 12/(x – 8) = -1

⇒ 4/(x-8) =-1

⇒ 4 = 8 – x

⇒ x = 4

Question 18. Find the value of x for which the points (x,-1), (2,1) and (4,5) are collinear.

Solution:

Let the given points are P (x,-1), Q (2,1) and R (4,5)

Also, given that the points P, q, and R are collinear, therefore 

the area of the triangle that they form must be zero.

Hence,

x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂) = 0

⇒ x(1 – 5) + 2(5 – (-1)) + 4(-1 – 1) = 0

⇒ -4x + 2(5 + 1) + (-2) = 0

⇒ -4x + 12 -8 = 0

⇒ -4x = -4

⇒ x = 1

Question 19. Find the angle between x-axis and the line joining the points (3,-1) and (4,-2).

Solution:

Slope of x axis m1 = 0

Slope of the line = [y₂ – y₁] / [x₂ – x₁]

m2 = [-2 – (-1)] / [4 – 3]

m2 = -1/1 = -1

Let us considered θ be the angle between x-axis and 

the line joining the points (3,-1) and (4,-2).

tanθ = [m1 – m2] / [1 + m1m2]

= [0 – (-1)] / [1 + (0)(-1)]

= -1

∴ θ = 3π/4

Question 20. By using concept of slope, show that the points (-2,-1), (4,0), (3,3), and (-3,2) are the vertices of a parallelogram.

Solution:

Let P (-2,-1), Q (4,0), R (3,3) and S (-3,2) be the vertices of the quadrilateral

We know

Slope of the line = [y₂ – y₁] / [x₂ – x₁]

Slope of the line PQ = [0 – (-1)] / [4 – (-2)]

= 1/6

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

= -3

Slope of the line RS = [2 – 3] / [-3 – 3]

= 1/6

Slope of the line RP = [2 – (-1)] / [-3 – (-2)]

= -3

We see that the slope of opposite side of the quadrilateral PQRS are equal.

Hence, the quadrilateral PQRS is a parallelogram.

Question 21. A quadrilateral has vertices (4,1), (1, 7), (-6, 0) and (-1, -9). Show that the mid-points of the sides of this quadrilateral form a parallelogram.

Solution:

Let P (4,1), Q (1, 7), R (-6, 0) and S (-1, -9) be the vertices of the quadrilateral

Let W, X, Y and Z be the mid points of PQ, QR, RS, and SP respectively

Using the mid point formula 

[(x₁ + x₂)/2, (y₁ + y₂)/2]

Mid point of PQ, 

W = [(4 + 1)/2, (1 + 7)/2] = (5/2, 4)

Mid point of QR, 

X = [(1 – 6)/2, (7 + 0)/2] = (-5/2, 7/2)

Mid point of RS, 

Y = [(-6 – 1)/2, (0 – 9)/2] = (-7/2, -9/2)

Mid point of SP, 

Z = [(-1 + 4)/2, (-9 + 1)/2] = (3/2, -4)

We know that diagonals of a parallelogram intersect at their mid points

Mid point of diagonal WY

= [{(5 – 7)/2}/2, {(4 – 9/2)/2}] = (-2/4, -1/4) = (-1/2, -1/4)

Mid point of diagonal XZ

= [{(-5 + 3)/2}/2, {(7 – 8/2)/2}] = (-2/4, -1/4) = (-1/2, -1/4)

Thus, the mid-points of the sides of this quadrilateral form a parallelogram.