Question

Vinay goes 150m due east and then 200m due north. How far is he from the starting point?

Answer 1

Answer:

250m

Step-by-step explanation:

Vinay goes 150m due east and then 200m due north.

You can say, Vinay at first goes 150 unit along positive X-axis and then he goes 200 unit along parallel to positive Y-axis. So his coordinate is (150,200). The distance from origin is $\sqrt{150^{2} +200^{2} } = \sqrt{62500} = 250$ unit

Therefore he goes 250m away from the starting point.

Answer 2

Given:

Vinay goes 150m due east and then 200m due north.

To Find:

How far is he from the starting point?

Solution:

We can find the distance from the starting point by either using a Pythagoras theorem or distance formula. Let the starting coordinates be (0,0) and it is given that

He moves 150 due east that means he has moved +150 in the x-axis, so the new coordinates will be (150,0), now

He moves 200m due north that means he has moved +200 in the y-axis, so the new coordinates will be (150,200)

Now the formula for distance is,

$d=\sqrt{(x_{1}-x_{2})^2+(y_{1}-y_{2})^2}$

Now putting the values as,

$d=\sqrt{150^2+200^2}\\=\sqrt{62500} \\=250$

Hence, the value of the distance is 250m.