Question

Two particles start from the point (2, -1), one moves 2 units along the line x + y = 1 and the other 5 units along the line x - 2y = 4. If the particles move towards increasing y, then their new positions are

• A. (2 -√2, √2 –1), (2√5 +2, √5 –1)
• A. (2√5 + 2, √5 – 1), (2 +√2, √2 +1)
• A. (2+√2, √2 + 1), (2√5 +2, √5 +1)
• A. None of these

Answer 1

(2 -√2, √2 –1), (2√5 +2, √5 –1)

Answer 2

The starting point is $P(2, -1)$.

For the first particle, the line is $x+y=1$. The slope is $m_1 = -1$. The angle with the positive x-axis is $135^\circ$. The new position $P_1$ is: $P_1 = (2 + 2\cos 135^\circ, -1 + 2\sin 135^\circ) = (2 + 2(-\frac{1}{\sqrt{2}}), -1 + 2(\frac{1}{\sqrt{2}})) = (2 - \sqrt{2}, -1 + \sqrt{2})$.

For the second particle, the line is $x-2y=4$. The slope is $m_2 = 1/2$. Let $\theta$ be the angle with the positive x-axis. Then $\tan\theta = 1/2$. From this, we can find $\cos\theta = \frac{2}{\sqrt{1^2+2^2}} = \frac{2}{\sqrt{5}}$ and $\sin\theta = \frac{1}{\sqrt{5}}$. The new position $P_2$ is: $P_2 = (2 + 5\cos\theta, -1 + 5\sin\theta) = (2 + 5(\frac{2}{\sqrt{5}}), -1 + 5(\frac{1}{\sqrt{5}})) = (2 + 2\sqrt{5}, -1 + \sqrt{5})$.

The new positions are $(2 - \sqrt{2}, -1 + \sqrt{2})$ and $(2 + 2\sqrt{5}, -1 + \sqrt{5})$.