Chapter 10: Straight Lines Exercise 10.1

Let us consider ‘m₁’ and ‘m’ to be the slope of the two given lines, such that m₁ = 2m

We know that if θ is the angle between the lines l1 and l2 with slope m₁ and m₂, then

1+2m² = -3m

2m² +1 +3m = 0

2m (m+1) + 1(m+1) = 0

(2m+1) (m+1)= 0

m = -1 or -1/2

If m = -1, then the slope of the lines are -1 and -2.

If m = -1/2, then the slope of the lines are -1/2 and -1.

Case 2:

2m² – 3m + 1 = 0

2m² – 2m – m + 1 = 0

2m (m – 1) – 1(m – 1) = 0

m = 1 or 1/2

If m = 1, then the slope of the lines are 1 and 2.

If m = 1/2, then the slope of the lines are 1/2 and 1.

∴ The slope of the lines are [-1 and -2] or [-1/2 and -1] or [1 and 2] or [1/2 and 1]

Given, the slope of the line is ‘m’.

The slope of the line passing through (x₁, y₁) and (h, k) is (k – y₁)/(h – x₁).

So,

(k – y₁)/(h – x₁)  = m

(k – y₁) = m (h – x₁)

Hence, proved.

Let us consider if the given points A (h, 0), B (a, b) and C (0, k) lie on a line.

Then, the slope of AB = slope of BC

(b – 0)/(a – h) = (k – b)/(0 – a)

Let us simplify, and we get

-ab = (k-b) (a-h)

-ab = ka- kh –ab +bh

ka +bh = kh

Divide both sides by kh, and we get

ka/kh + bh/kh = kh/kh

a/h + b/k = 1

Hence, proved.

We know that line AB passes through points A (1985, 92) and B (1995, 97).

Its slope will be (97 – 92)/(1995 – 1985) = 5/10 = 1/2

Let ‘y’ be the population in the year 2010. Then, according to the given graph, AB must pass through point C (2010, y).

So now, slope of AB = slope of BC

15/2 = y – 97

y = 7.5 + 97 = 104.5

∴ The slope of line AB is 1/2, while in the year 2010, the population will be 104.5 crores.