Question:

The point (4, 1) undergoes the following three transformations successively I. Reflection about the line y = x. II. Transformation through a distance 2 units along the positive direction of X-axis. III. Rotation through an angle $\frac{\pi}{4}$ about the origin in the counter clockwise direction. Then, the final position of the point is given by the coordinates

Updated On: Jul 7, 2022

$\Bigg(\frac{1}{\sqrt{2}},\frac{7}{\sqrt{2}}\Bigg)$
$(-\sqrt{2}, 7\sqrt{2})$
$\Bigg(-\frac{1}{\sqrt{2}},\frac{7}{\sqrt{2}}\Bigg)$
$(\sqrt{2}, 7\sqrt{2})$

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The Correct Option is C

Solution and Explanation

Let B, C, D be the position of the point A (4,1) after the three operations I, II and III, respectively. Then, B is (1, 4), C(1 + 2,4) i.e. (3, 4). The point D is obtained from C by rotating the coordinate axes through an angle $\pi/4$ in anti-clockwise direction. Therefore, the coordinates of D are given by $X =3cos \frac{\pi}{2}-4 sin \frac{\pi}{4}=-\frac{-1}{\sqrt{2}}$ and $Y=3 sin \frac{\pi}{4}+4 cos\frac{\pi}{4}=\frac{7}{\sqrt{2}}$ $\therefore $ Coordinates of D are $\Bigg(-\frac{1}{\sqrt{2}},\frac{7}{\sqrt{2}}\Bigg).$

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Concepts Used:

Straight lines

A straight line is a line having the shortest distance between two points.

A straight line can be represented as an equation in various forms, as show in the image below:

The following are the many forms of the equation of the line that are presented in straight line-

1. Slope – Point Form

Assume P0(x0, y0) is a fixed point on a non-vertical line L with m as its slope. If P (x, y) is an arbitrary point on L, then the point (x, y) lies on the line with slope m through the fixed point (x0, y0) if and only if its coordinates fulfil the equation below.

y – y₀ = m (x – x₀)

2. Two – Point Form

Let's look at the line. L crosses between two places. P₁(x₁, y₁) and P₂(x₂, y₂) are general points on L, while P (x, y) is a general point on L. As a result, the three points P₁, P₂, and P are collinear, and it becomes

The slope of P₂P = The slope of P₁P₂, i.e.

$\frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1}$

Hence, the equation becomes:

y - y₁ =$ \frac{y_2-y_1}{x_2-x_1} (x-x1)$

3. Slope-Intercept Form

Assume that a line L with slope m intersects the y-axis at a distance c from the origin, and that the distance c is referred to as the line L's y-intercept. As a result, the coordinates of the spot on the y-axis where the line intersects are (0, c). As a result, the slope of the line L is m, and it passes through a fixed point (0, c). The equation of the line L thus obtained from the slope – point form is given by

y – c =m( x - 0 )

As a result, the point (x, y) on the line with slope m and y-intercept c lies on the line, if and only if

y = m x +c