First slide

Equation of line in Straight lines

Question

The point P (1, 1) is translated parallel to $2 x = y$ in the first quadrant through a unit distance. The coordinates of the new position of P are

Difficult

A

$(1 \pm \frac{2}{\sqrt{5}}, 1 \pm \frac{1}{\sqrt{5}})$
B

$(1 \pm \frac{1}{\sqrt{5}}, 1 \pm \frac{2}{\sqrt{5}})$
C

$(\frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}})$
D

$(\frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}})$

Solution

$\tan θ = 2$
$∴ \sin θ = \frac{2}{\sqrt{5}}, \cos θ = \frac{1}{\sqrt{5}}$
$L e t n e w c o - o r d i n a t e i s (x_{1}, y_{1})$
$∴ \frac{x_{1} - 1}{\cos θ} = \frac{y_{1} - 1}{\sin θ} = \pm 1$
$∴ x_{1} = 1 \pm \cos θ, y_{1} = 1 \pm \sin θ$
$x_{1} = 1 \pm \frac{1}{\sqrt{5}}, y_{1} = 1 \pm \frac{2}{\sqrt{5}}$
$H e n c e, n e w p o s i t i o n o f P a r e (1 \pm \frac{1}{\sqrt{5}}, 1 \pm \frac{2}{\sqrt{5}})$

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