Q. 375.0( 1 Vote )

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-axisThen the final coordinates of the point areA. (4, 3)B. (3, 4)C. (1, 4)D.

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)

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