Question

cartesian plane
using cartesian coordinates be mark a point on a graph by how far along and how far far up it is the left right horizontal direction is commonly called x-axis dubbed on vertical direction is commonly called y-axis when we include negative values and x and y axis divides the space up to you to 4 pieces read the following passage and answer the questions that follow using the above information in a classroom for students Seeta ,Geeta ,Rita and Anita are sitting at A(3,4) B(6,7) C(9,4) D(6,1) respectively is a new student Anjali join the class

Answer 1

Answer:

Anjali's co-ordinates : (6,4)

Distance b/w Sita and Anita is 3√2units.

Two students are equidistant from Gita are Sita and Rita

The geometrical figure formed after joining the ABCD is a Square

The distance between Sita and Rita is 6 units

HOPE IT HELPS

Answer 2

SOLUTION

GIVEN

Seeta ,Geeta ,Rita and Anita are sitting at A(3,4) B(6,7) C(9,4) D(6,1) respectively. A new student Anjali join the class

TO DETERMINE

(i) Teacher tells Anjali to sit in the middle of the four students. Find the coordinates of the position where she can sit.

(ii) Calculate the distance between Sita and Anita.

(iii) Which two students are equidistant from Gita.

EVALUATION

Here it is given that Seeta ,Geeta ,Rita and Anita are sitting at A(3,4) B(6,7) C(9,4) D(6,1) respectively.

Now a new student Anjali join the class

(i) Teacher tells Anjali to sit in the middle of the four students.

So the coordinates of the position where she can sit is the middle point of AC ( or BD )

Hence the coordinates of the position where she can sit

= Middle point of AC

$\displaystyle \sf{= \bigg( \frac{3 + 9}{2} \: , \: \frac{4 + 4}{2} \bigg) }$

$\displaystyle \sf{= \bigg( \frac{12}{2} \: , \: \frac{8}{2} \bigg) }$

$\displaystyle \sf{= ( 6 \: , \: 4 ) }$

(ii) The distance between Sita and Anita

= The distance between A & D

$\sf{ = \sqrt{ {(3 - 6)}^{2} + {(4 - 1)}^{2} \: \: } \: \: unit}$

$\sf{ = \sqrt{ {3}^{2} + {3}^{2} \: \: } \: \: unit}$

$\sf{ = \sqrt{9 + 9 \: \: } \: \: unit}$

$\sf{ = \sqrt{18 \: \: } \: \: unit}$

$\sf{ = 3\sqrt{2 } \: \: unit}$

(iii) The distance between A & B

$\sf{ = \sqrt{ {(3 - 6)}^{2} + {(4 - 7)}^{2} } \: \: unit}$

$\sf{ = \sqrt{ {3}^{2} + {3}^{2} } \: \: unit}$

$\sf{ = \sqrt{18} \: \: unit}$

$\sf{ =3 \sqrt{ 2 } \: \: unit}$

Distance between B & C

$\sf{ = \sqrt{ {(9 - 6)}^{2} + {(4 - 7)}^{2} } \: \: unit}$

$\sf{ = \sqrt{ {(3 )}^{2} + {(3)}^{2} } \: \: unit}$

$\sf{ =3 \sqrt{2 } \: \: unit}$

Distance between B & D

$\sf{ = \sqrt{ {(6- 6)}^{2} + {(1 - 7)}^{2} } \: \: unit}$

$\sf{ = \sqrt{ {(0)}^{2} + {(6)}^{2} } \: \: unit}$

$\sf{ = 6 \: \: unit}$

Hence A & C are equidistant from B

Hence Seeta and Rita is equidistant from Geeta

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