First slide

Equation of line in Straight lines

Question

Point P $(2, 4)$ is translated through a distance $3 \sqrt{2}$ units measured parallel to the line $y - x - 1 = 0$ , in the direction of decreasing ordinates, to reach at $Q .$ $I f R$ is the image ofQ unth respect to the line $y - x - 1 = 0$ , then coordinates of $R$ are

Moderate

A

(-1, 1)
B

(5, 7)
C

(6, 6)
D

(0, 0)

Solution

The equation of a line through P (2, 4) and parallel to y - x - 1 = 0 is
$\frac{x - 2}{\cos π / 4} = \frac{y - 4}{\sin π / 4}$
The coordinates of Q are given by
$\frac{x - 2}{\cos π / 4} = \frac{y - 4}{\sin π / 4} = - 3 \sqrt{2} \Rightarrow x = - 1, y = 1$
Thus, the coordinates of Q are $(- 1, 1) . R$ s the image of Q with respect to the line $y - x - 1 = 0 .$
Therefore, coordinates of R are given by
$\begin{aligned} \frac{x + 1}{- 1} = \frac{y - 1}{1} = - 2 \frac{(1 + 1 - 1)}{(- 1)^{2} + (1)^{2}} \\ \frac{x + 1}{- 1} = \frac{y - 1}{1} = - 1 \Rightarrow x = 0, y = 0 \end{aligned}$
Hence, the coordinates of R are (0, 0).

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