

InterviewSolution
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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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