CAT 2023 Slot 2 QA Question 19

The question is about finding the area of a quadrilateral bounded by the Y-axis, the line x=5, and the lines given by the equation |x - y| - |x - 5| = 2. Hmm, modulus equations can sometimes be tricky because they involve absolute values, which can split into different cases. when dealing with absolute values, it's often helpful to consider different regions based on the critical points where the expressions inside the absolute values change their signs.

First, We have the equation again to make sure I have it right:

|x - y| - |x - 5| = 2.

We find the area of the quadrilateral formed by this equation along with the Y-axis (x=0) and the line x=5. So, the boundaries are x=0, x=5, and the two lines from the modulus equation.

I think the first step is to analyze the equation |x - y| - |x - 5| = 2. Let me try to solve for y in terms of x so that I can understand the lines involved.

Let A = |x - y| and B = |x - 5|. So, the equation becomes A - B = 2, which implies A = B + 2. Therefore, |x - y| = |x - 5| + 2.

Now, since absolute values are always non-negative, |x - 5| + 2 is always greater than or equal to 2. So, |x - y| must be at least 2.

To solve |x - y| = |x - 5| + 2, I can consider different cases based on the value of x. The critical points occur where the expressions inside the absolute values change their signs, which happens when x - y = 0 and x - 5 = 0. So, x = y and x = 5 are the critical points.

But since we're dealing with a modulus equation involving both x and y, it's a bit more complex. Maybe I can square both sides to eliminate the absolute values, but I'm not sure if that's the best approach here. Alternatively, I can consider different regions based on the value of x relative to 5.

Let me think about the regions:

In this case, |x - 5| = x - 5. So, the equation becomes |x - y| = (x - 5) + 2 = x - 3.

Now, |x - y| = x - 3 implies that either x - y = x - 3 or x - y = -(x - 3).

Let's solve both:

So, for x >= 5, the equation represents two lines: y = 3 and y = 2x - 3.

Here, |x - 5| = 5 - x. So, the equation becomes |x - y| = (5 - x) + 2 = 7 - x.

Again, |x - y| = 7 - x implies two cases:

So, for x < 5, the equation represents two lines: y = 7 and y = 2x - 7.

Wait, but I need to make sure that these solutions are valid in their respective regions. For example, when x >= 5, y = 3 and y = 2x - 3. Let me check if these lines are valid beyond x=5.

Similarly, for x < 5, y = 7 and y = 2x - 7. if these lines make sense in their regions.

Now, I can plot these lines to visualize the quadrilateral. The boundaries are x=0 (Y-axis), x=5, and the four lines from the modulus equation.

But wait, the equation |x - y| - |x - 5| = 2 is a single equation, so it's actually two lines in each region. So, the overall figure is bounded by x=0, x=5, and the four lines from the modulus equation. But since it's a quadrilateral, it must have four sides, so perhaps the lines intersect in such a way that they form a four-sided figure.

Let me try to find the points of intersection of these lines with each other and with x=0 and x=5.

First, let's consider the lines for x >= 5:

And for x < 5:

Now, let's find where these lines intersect the boundaries x=0 and x=5.

Starting with x=5:

For x=5, in the region x >=5, the lines are y=3 and y=2*5 -3=7. So, at x=5, the points are (5,3) and (5,7).

Similarly, for x=0, in the region x <5, the lines are y=7 and y=2*0 -7=-7. So, at x=0, the points are (0,7) and (0,-7).

Wait, but the quadrilateral is bounded by Y-axis (x=0), x=5, and the lines from the modulus equation. So, the four vertices of the quadrilateral would be the intersection points of these lines with x=0 and x=5, and where the lines from the modulus equation intersect each other.

Let me find the intersection points:

Set 3 = 2x -3 => 2x=6 => x=3. So, the point is (3,3).

Set 7 = 2x -7 => 2x=14 => x=7. But x=7 is beyond x=5, which is our boundary. So, this intersection is outside our region of interest.

Set 3 = 2x -7 => 2x=10 =>