(a) Let P(x₁, y₁) be any point on the given line x + 5y = 13

∴ x₁ + 5y₁ = 13

⇒ 5y₁ = 13 – x₁ …(i)

Distance of the point P(x₁,y₁) from the equation 12x – 5y + 26 = 0

[given d = 2]

[from eq. (i)]

⇒ 2 = |x₁ + 1|

⇒ 2 = ±(x₁ + 1)

either x₁ + 1 = 2 or -(x₁ + 1) = 2

x₁ = 1 …(ii) or x₁ + 1 = -2

or x₁ = -3 …(iii)

Putting the value of x₁ = 1 in eq. (i), we get

5y₁ = 13 – 1

⇒ 5y₁ = 12

Putting the value of x₁ = -3 in eq. (i), we get

5y₁ = 13 – (-3)

⇒ 5y₁ = 13 + 3

⇒ 5y₁ = 16

Hence, the required points on the given line are

Hence, (a) ↔ (iii)

(b) Let P(x₁,y₁) be any point lying in the equation x + y = 4

∴ x₁ + y₁ = 4 …(i)

Distance of the point P(x₁,y₁) from the equation 4x + 3y = 10

[given]

⇒ 4x₁ + 3y₁ – 10 = ±5

either 4x₁ + 3y₁ – 10 = 5 or 4x₁ + 3y₁ – 10 = -5

4x₁ + 3y₁ = 5 + 10 or 4x₁ + 3y₁ = -5 + 10

4x₁ + 3y₁ = 15 …(ii) or 4x₁ + 3y₁ = 5 …(iii)

From eq. (i), we have y₁ = 4 – x₁ …(iv)

Putting the value of y₁ in eq. (ii), we get

4x₁ + 3(4 – x₁) = 15

⇒ 4x₁ + 12 – 3x₁ = 15

⇒ x₁ = 15 – 12

⇒ x₁ = 3

Putting the value of x₁ in eq. (iv), we get

y₁ = 4 – 3

⇒ y₁ = 1

Putting the value of y₁ = 4 – x₁ in eq. (iii), we get

4x₁ + 3(4 – x₁) = 5

⇒ 4x₁ + 12 – 3x₁ = 5

⇒ x₁ = 5 – 12

⇒ x₁ = - 7

Putting the value of x₁ in eq. (iv), we get

y₁ = 4 – (-7)

⇒ y₁ = 4 + 7

⇒ y₁ = 11

Hence, the required points on the given line are (3,1) and (-7,11)

Hence, (b) ↔ (i)

(c) Given that AP = PQ = QB

and given points are A (-2, 5) and B (3, 1)

Firstly, we find the slope of the line joining the points (-2, 5) and (3, 1)

Now, equation of line passing through the point (-2, 5)

y – y₁ = m(x – x₁)

⇒ 5y – 25 = -4(x + 2)

⇒ 5y – 25 = -4x – 8

⇒ 4x + 5y – 25 + 8 = 0

⇒ 4x + 5y – 17 = 0

Let P(x₁, y₁) and Q(x₂, y₂) be any two points on the AB

P(x₁, y₁) divides the line AB in the ratio 1:2

So, the coordinates of P(x₁, y₁) is

Now, Q(x₂, y₂) is the midpoint of PB

Hence, the coordinates of Q(x₂, y₂) is

Hence, (c) ↔ (ii)