Question

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Answer 1

❖ Question :–

Q) A policeman and a thief are equidistant from a jewel box. Upon considering jewel box as origin, the position of Policeman is (0,5) . If the ordinate of the position of thief is zero, then write the coordinates of the position of thief.

❖ Answer :–

» Consider the attachment.

» In the attachment,

• P represents the Policeman.

• J represents the Jewel Box.

• T represents the Thief.

$\frak{\underline{\dag\; Given:}}$

• PJ = TJ (Since they are equidistant)

• Coordinates of J = (0,0)

• Coordinates of P = (0,5)

• Ordinate of T = 0

$\frak{\underline{\dag\;To\:Find:}}$

• The Coordinates of point T.

$\frak{\underline{\dag\;Solution:}}$

» Since the points are equidistant so, the distance PJ would be equal to the distance TJ.

» We can find the distance between two points using the Distance Formula.

$\odot \: \boxed{ \tt d = \sqrt{(x _{2} - x _{1} ) {}^{2} + {(y _{2} - y _{1}) }^{2} } }$

◐ Finding the distance PJ ◑

For PJ,

• $\sf x_1=0, y_1=5$ &

• $\sf x_2=0, y_2= 0$

So,

$\colon\rarr\sf\: PJ=\sqrt{(0-0)^2+(0-5)^2}\\\\\sf\colon\rarr\: PJ=\sqrt{0+(-5)^2}\\\\\sf\colon\rarr\: PJ=\sqrt{25}\\\\\colon\dashrightarrow\boxed{\pink{\frak{PJ=5\:units}}}$

◐ Finding the distance TJ ◑

For TJ,

• $\sf x_1=a, y_1=0$

• $\sf x_2=0, y_2=0$

So,

$\colon\sf\rarr\: TJ=\sqrt{(0-a)^2+(0-0)^2}\\\\\sf\colon\rarr\: TJ = \sqrt{(-a)^2+0}\\\\\sf\colon\rarr\:TJ=\sqrt{a^2}\\\\\colon\dashrightarrow\boxed{\pink{\frak{TJ=a\:units}}}$

◐ Equating PJ & TJ ◑

$\colon\implies\sf\: PJ=TJ \\\\\sf\colon\dashrightarrow\boxed{\red{\frak{a=5}}}$

• T = ( a,0 ) = ( 5,0 )

$\therefore\;\underline{\sf Coordinates\:of\:the\:position\:of\:the\:Thief=\bf{(5, 0) .}}$

_____________________________

Answer 2

Answer:

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Step-by-step explanation:

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