First slide

Equation of line in Straight lines

Question

The point (2, l) is shifted by $3 \sqrt{2}$ units parallel $x + y = 1$ the line in the direction of increasing ordinate, to reach Q The image of C2 in the line $x + y = 1$ , is

Moderate

A

(5, -- 2)
B

(--3,2)
C

(5, 4)
D

(-1, 4)

Solution

The equation of a line passing through (2, 1) and parallel to x + y = 1 is
$\frac{x - 2}{\cos 3 π / 4} = \frac{y - 1}{\sin π / 4} or, \frac{x - 1}{- 1 / \sqrt{2}} = \frac{y - 1}{1 / \sqrt{2}}$
The coordinates of points on this line which are at a distance $3 \sqrt{2}$ from (2, 1) are given by
$\frac{x - 2}{- 1 / \sqrt{2}} = \frac{y - 1}{1 / \sqrt{2}} = \pm 3 \sqrt{2} \Rightarrow x = 2 + 3, y = 1 \pm 3$
But, the ordinate of Q is more than that of P ( 2, 1 ). So, the coordinates of Q are (- 1, 4). The image of Q (- 1, 4) in the line x + y = 1 is given by
$\begin{aligned} \frac{x + 1}{1} = \frac{y - 4}{1} = - 2 (\frac{- 1 + 4 - 1}{1^{2} + 1^{2}}) \\ \frac{x + 1}{1} = \frac{y - 4}{1} = - 2 \Rightarrow x = - 3, y = 2 \end{aligned}$
Thus, the coordinates of the required point are (- 3, 2)

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