Question

find the area enclosed by the curve y=x with x-axis and x=1 using determinants​

Answer 1

Given :   area enclosed by the curve y=x with x-axis and x=1  

To Find : Area

Solution:

y = x

x -axis => y = 0

x = 1

y = x   &  y = 0

=> (0 ,0)  is the point

y = x   and x = 1

=> ( 1 , 1)  is the point

y = 0  , x = 1

=> (1 , 0) is the point

(0 , 0) , (0 , 1) , ( 1, 1)

$A=\dfrac{1}{2}\begin{vmatrix} 0&0&1\\0&1&1\\1&1&1\end{vmatrix}$

=> A = (1/2) | 0 ( 1 - 1)  - 0(0 - 1)  + 1 (0 - 1) |

=> A = (1/2) | - 1|

=> A = 1/2

area enclosed by the curve y=x with x-axis and x=1  is 1/2 sq units

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Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

If the vertices of a triangle ABC is

$(x_1, y_1), (x_2, y_2), (x_3, y_3) \: \: then$

Area of the triangle ABC is

$= \frac{1}{2} \displaystyle\begin{vmatrix} x_1 & y_1 & 1\\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \: \: \sf{sq \: unit}$

TO DETERMINE

The area enclosed by the curve y=x with x-axis and x=1 using determinants

CALCULATION

We first find the points of intersection of the line with coordinate axis

For the point of intersection of the line y = x and x = 1

$\sf{we \: put \: x = 1 \: in \: y = x }$

$\sf{Which \: gives \: y = 1}$

So the point of intersection is ( 1, 1 )

So the vertices of the triangle OAB are

$( 0, 0), (1, 0), (1,1)$

The shaded region is bounded by the line y = x, x axis and x = 1

So by the determinant method the area of the Triangle OAB is

$= \frac{1}{2} \displaystyle\begin{vmatrix} 0 & 0 & 1\\ 1& 0 & 1 \\ 1 & 1 & 1 \end{vmatrix} \: \: \sf{sq \: unit}$

$= \displaystyle \frac{1}{2} \: \: \: \sf{sq \: unit}$

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