Satpal walks 23 km from a place P- towards east and then from there 157 km towards west- Where will he be now from P-

R.E.F. Image.In POB, PB = √((2/3)^2+(12/7)^2) = √(4/9+144/49) = √(196+1296/441) = √(1492/441) = PB = 38.62/21km. Satpal is now 38.62/21 ...

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Standard X

Mathematics

Distance between Two Points on the Same Coordinate Axes

Satpal walks ...

Question

Satpal walks $\frac{2}{3} k m$ from a place $P$ , towards east and then from there $1 \frac{5}{7} k m$ towards west. Where will he be now from $P$ ?

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Solution

R.E.F. Image.
In $△ P O B,$
$P B = \sqrt{(\frac{2}{3})^{2} + (\frac{12}{7})^{2}} = \sqrt{\frac{4}{9} + \frac{144}{49}}$
$= \sqrt{\frac{196 + 1296}{441}} = \sqrt{\frac{1492}{441}}$
$= P B = \frac{38.62}{21} k m .$
Satpal is now $\frac{38.62}{21}$ kms north - east from P.

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Satpal walks 23 km from a place P, towards east and then from there 157 km towards west. Where will he be now from P?

R.E.F. Image.In △POB,PB=√(23)2+(127)2=√49+14449=√196+1296441=√1492441=PB=38.6221km.Satpal is now 38.6221 kms north - east from P.

Satpal walks $\frac{2}{3} k m$ from a place $P$ , towards east and then from there $1 \frac{5}{7} k m$ towards west. Where will he be now from $P$ ?

R.E.F. Image.
In $△ P O B,$
$P B = \sqrt{(\frac{2}{3})^{2} + (\frac{12}{7})^{2}} = \sqrt{\frac{4}{9} + \frac{144}{49}}$
$= \sqrt{\frac{196 + 1296}{441}} = \sqrt{\frac{1492}{441}}$
$= P B = \frac{38.62}{21} k m .$
Satpal is now $\frac{38.62}{21}$ kms north - east from P.