# CAT Practice : Coordinate Geometry

For a given equation, how do we construct and visualize the graph? We tackle this important idea here.

## Area under curve

Q.5: What is the area enclosed in the region defined by y = |x – 1| + 2, line x = 1, X–axis and Y–axis?
1. 5 sq units
2. 2.5 sq units
3. 10 sq units
4. 7 sq units

Choice B. 2.5 sq units

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## Detailed Solution

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 = ${1 \over 2}$ × height × sum of the parallel sides = ${1 \over 2}$ × 1 × (3 + 2) = ${5 \over 2}$ sq units.

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## More questions from Coordinate Geometry

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.