1.

The area enclosed within the curve |x| + |y| = 1 is:1. 2 sq. unit2. 3 sq. unit3. 1 sq. unit4. 5 sq. unit

Answer» Correct Answer - Option 1 : 2 sq. unit

Concept:

1. The intercept form of the line is  \({\rm{\;}}\frac{{\rm{x}}}{{\rm{a}}} + \frac{{\rm{y}}}{{\rm{b}}} = 1\) 

Where a is the x-intercept and b is the y-intercept.

2. Consider a square of side 'a'.

Area of square = a2

Diagonal of square = √2 a

Pythagoras theorem: It states that In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides.

Distance formula: Let  A = (x1, y1) and  B = (x2, y2) be any two points.

Then Distance between A And B is given by the distance formula.

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

Calculation:

Given curve is 

|x| + |y| = 1

⇒ ± x ± y = 1 

On comparing the above equation with the intercept form of line 

\({\rm{\;}}\frac{{\rm{x}}}{{\rm{a}}} + \frac{{\rm{y}}}{{\rm{b}}} = 1\)

We can see that, x & y-intercept are A(1, 0), B(0, 1), C(-1, 0) & D(0, -1)

Clearly ABCD represent square.

Using Pythagoras theorem,

The length of each side = Distance b/w AB or BC or CD or DA

⇒ \(a\ =\ \sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}\)

\(a \ =\ \sqrt{1^1\ +\ 1^2}\ =\ √{2}\)

Therefore, the area of square ABCD

A = (√2)2 = 2 sq. unit.

Hence, option 1 is correct.



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