
A policeman and a thief are equidistant from the jewel box, upon considering jewel box as origin, the position of the policeman is ( 0 , 5 ). If the ordinate of the position of the thief is zero. Then write the coordinates of the position of the thief.
Answer
556.5k+ views
Hint:
We are given that the position of the jewel box is at the origin and the coordinates of the police man to be (0 , 5). We are given that the ordinate of the position of the thief is 0. So let the abscissa of the position of the thief be x.And the coordinate of the position of the thief be Q(x,0). As the police and the thief are equidistant from the jewel box . we get OP = OQ and using the distance formula $\sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $we need to find the value of OP and OQ and equating them we get the value of x.
Complete step by step solution:
We are given that the position of the jewel box is at the origin. Let the position of the jewel box be O (0,0). Now let the position of the police man be P. And we are given the coordinates of the police man to be (0 , 5). Therefore P = (0,5). And we are given that the ordinate of the position of the thief is 0. Ordinate of point is nothing other than its y coordinate and the x coordinate is known as the abscissa. So, let the abscissa of the position of the thief be x
And the coordinate of the position of the thief be Q(x,0)
We are given that the police and the thief are equidistant from the jewel box
That is OP = OQ
Now we can use the distance formula $\sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $ to find OP and OQ
Distance between O(0 , 0) and P(0 , 5)
$
\Rightarrow OP = \sqrt {{{(0 - 0)}^2} + {{(5 - 0)}^2}} \\
\Rightarrow OP = \sqrt {{5^2}} = 5units \\
$
Distance between O(0 , 0) and Q(x , 0)
$
\Rightarrow OQ = \sqrt {{{(x - 0)}^2} + {{(0 - 0)}^2}} \\
\Rightarrow OQ = \sqrt {{x^2}} = xunits \\
$
We know that OP = OQ
$ \Rightarrow 5 = x$
Therefore we get the value of x to be 5.
Hence the coordinates of the position of the thief is (5 , 0).
Note:
If the police , thief and the jewel box are in a straight line. As we are given that the police and the thief are equidistant from the jewel box. Then the coordinates of the jewel box is the midpoint of the line joining the coordinates of the police and thief.
We are given that the position of the jewel box is at the origin and the coordinates of the police man to be (0 , 5). We are given that the ordinate of the position of the thief is 0. So let the abscissa of the position of the thief be x.And the coordinate of the position of the thief be Q(x,0). As the police and the thief are equidistant from the jewel box . we get OP = OQ and using the distance formula $\sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $we need to find the value of OP and OQ and equating them we get the value of x.
Complete step by step solution:
We are given that the position of the jewel box is at the origin. Let the position of the jewel box be O (0,0). Now let the position of the police man be P. And we are given the coordinates of the police man to be (0 , 5). Therefore P = (0,5). And we are given that the ordinate of the position of the thief is 0. Ordinate of point is nothing other than its y coordinate and the x coordinate is known as the abscissa. So, let the abscissa of the position of the thief be x
And the coordinate of the position of the thief be Q(x,0)
We are given that the police and the thief are equidistant from the jewel box
That is OP = OQ
Now we can use the distance formula $\sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $ to find OP and OQ
Distance between O(0 , 0) and P(0 , 5)
$
\Rightarrow OP = \sqrt {{{(0 - 0)}^2} + {{(5 - 0)}^2}} \\
\Rightarrow OP = \sqrt {{5^2}} = 5units \\
$
Distance between O(0 , 0) and Q(x , 0)
$
\Rightarrow OQ = \sqrt {{{(x - 0)}^2} + {{(0 - 0)}^2}} \\
\Rightarrow OQ = \sqrt {{x^2}} = xunits \\
$
We know that OP = OQ
$ \Rightarrow 5 = x$
Therefore we get the value of x to be 5.
Hence the coordinates of the position of the thief is (5 , 0).
Note:
If the police , thief and the jewel box are in a straight line. As we are given that the police and the thief are equidistant from the jewel box. Then the coordinates of the jewel box is the midpoint of the line joining the coordinates of the police and thief.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

