The point (4, 1) undergoes the following two successive transformations:

(i)Reflection about the line y = x

(ii)Translation through a distance 2 units along the positive x-axis

Then the final coordinates of the point are

Let Q(x, y) be the reflection of P(4, 1) about the line y = x, then midpoint of PQ

which lies on y = x

⇒ 4 + x = 1 + y

⇒ x – y + 3 = 0 …(i)

Now, we find the slope of given equation y = x

Since, this equation is in y = mx + b form.

So, the slope = m = 1

Slope of PQ =

Since, PQ is perpendicular to y = x

And we know that, when two lines are perpendicular then

m₁ m₂ = -1

⇒ y – 1 = - (x – 4)

⇒ y – 1 = - x + 4

⇒ x + y – 5 = 0 …(ii)

On adding eq. (i) and (ii), we get

x – y + 3 + x + y – 5 = 0

⇒ 2x – 2 = 0

⇒ x – 1 = 0

⇒ x = 1

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

1 – y + 3 = 0

⇒ -y + 4 = 0

⇒ y = 4

Given that translation through a distance 2 units along the positive x-axis

∴ The point after translation is (1 + 2, 4) = (3, 4)

Hence, the correct option is (b)