The question is from CAT Coordinate Geometry. This is about area under curve. We need to find out the area of the region which is enclosed by given lines. Take Geometry, add one unit of algebra; take a diagram, explain it with x's and y's. For the purists, it is geometry without the romance, for the pragmatists it is Geometry with expanded scope. It is important to cover ideas from Coordinate Geometry in your CAT Preparation

Question 5: What is the area enclosed in the region defined by y = |x – 1| + 2, line x = 1, X–axis and Y–axis?

- 5 sq units
- 2.5 sq units
- 10 sq units
- 7 sq units

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Let us first draw y = |x|.

Now, y = |x – 1| is just a shift of ‘1’ unit to the right along the x–axis.

Now, y = |x – 1| + 2 is a shift of ‘2’ units to the top along the y–axis.

Now, let us complete the diagram by drawing the line x = 1 also.

The required area is essentially the area of the shaded region.

The point of intersection of y = |x – 1| + 2 where the y–axis can be found by substituting x = 0 in the equation.

Thus we get y = 3.

Required area = Area of the trapezium formed by the points (0, 0), (1, 0), (1, 2) and (0, 3).

Area of a trapezium = \\frac{1}{2}\\) * height * sum of the parallel sides = \\frac{1}{2}\\) * 1 * (3 + 2) = \\frac{5}{2}\\) sq units.

The question is **"What is the area enclosed in the region defined by y = |x – 1| + 2, line x = 1, X–axis and Y–axis?"**

Choice B is the correct answer.

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