Question

Given figure shows the arrangement of desks in a classroom. Ashima, Bharti and Camella are seated
at A(3, 1), B(6. 4) and C(8. 6) respectively.
(1) Do you think they are seated in a line ? Give reasons for your answer.
Row
7
8
9
10
1
2
3
4 5 6
Columns
(ii) Which mathematical concept is used in the above problem?​

Answer 1

( i )They can be seated in a line. Because the points are collinear.

             AB + BC = AC

        3$\sqrt{2}$ + 2$\sqrt{2}$  =  5$\sqrt{2}$.  

( ii ) Mathematical concept used in the given problem is co-ordinate geometry.

Given

Ashima , Bharti and Camella are seated at distance of A ( 3, 1 ), B ( 6, 4 ) and C ( 8, 6 )

( i ) To find whether they are seated in a line :

To find they are seated in a line use the concept of collinear.  

                                   AB + BC = AC

Use distance formula to find the distance between two points.

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

                               A ( 3, 1 )           B ( 6, 4 )

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

            $x_{1}$ = 3              $x_2$ = 6          $y_1$ = 1        $y_{2}$ = 4          

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

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

                        = $\sqrt{2(3)^2}$  

                        = $\sqrt{2}$ × 3

                  AB = 3$\sqrt{2}$

                               B ( 6, 4 )             C ( 8, 6 )

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

               $x_{1}$ = 6             $x_2$ = 8          $y_1$ = 4        $y_{2}$ = 6

                        = $\sqrt{(8-6)^2 + (6-4)^2}$

                         = $\sqrt{(2)^2 + (2)^2}$

                        = $\sqrt{2(2)^2}$

                        = $\sqrt{2}$ × 2

                  BC = 2$\sqrt{2}$

                                 A ( 3, 1 )               C ( 8, 6 )

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

                 $x_{1}$ =3              $x_2$ = 8          $y_1$ = 1        $y_{2}$ = 6  

                         = $\sqrt{(8-3)^2+(6-4)^2}$

                         = $\sqrt{(5)^2+(5)^2}$

                         = $\sqrt{2(5)^2}$

                         = $\sqrt{2}$  × 5

                   AC = 5$\sqrt{2}$

Now, find whether the resultant is collinear

       AB = 3$\sqrt{2}$         BC = 2$\sqrt{2}$          AC = 5$\sqrt{2}$

                         AB + BC = AC

                         3$\sqrt{2}$ + 2$\sqrt{2}$  =  5$\sqrt{2}$.

Hence, it is a collinear. So, they can be seated in a line.

( ii ) Mathematical concept used in the above problem is co-ordinate geometry.

To learn more...

brainly.in/question/1097833

                         

Answer 2

Answer:

Step-by-step explanation:

Hey friend here is your answer

There are two ways to solve this problem:-

1)By using distance formula.

2)By using area of triangle formula.

Since our friend has already posted an answer using distance formula ,I would use area of triangle formula.

If the points are collinear then their area would be equal to zero.

A(3,1) B(6,4) ,C(8,6)

In the formula 1/2 can be transposed to the right.

So formula becomes,

x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2) = 0

=>3(4 - 6) + 6(6 - 1) + 8(1 - 4) = 0

=>3(-2) + 6(5) + 8(-3) = 0

=> - 6 + 30 - 24 = 0

Since the area is equal to zero,the points are collinear. So the students are seated on the same line.

ii) The mathematical concept used in the problem is co-ordinate geometry.

Thank u for asking the question

PLZZ mark as the brainliest.