First of all, we will learn how to plot the graph of this. Once, we know how to plot, we can find out the area and perimeter orally without plotting the graph.
If there is a single modulus involved in expression, we get two equations. But if two mods are involved, we will get four equations under different conditions. In this case also, we will get 4 equations.
1st equation: x + y = 5 if x ≥ 0 and y ≥ 0.
2nd equation: x – y = 5 if x ≥ 0 and y < 0.
3rd equation: –x + y = 5 if x < 0 and y ≥ 0.
4th equation: –x – y = 5 if x < 0 and y < 0.
So, we get four different equations with different conditions. So we need to plot these lines and use the conditions in which quadrant x and y is positive or negative.
If we merge all four graphs, we will get a square or a rhombus or a diamond. If we did this much of hard work in plotting, we can easily plot any graph of this type orally now if point changes from 5 to any other point.
There are two ways of solving these types of questions.
Algebraic method:
When x = 0, y can take two value which is ±5. (2 solutions).
When x = ±1, y can take two values which is ±4. [Total of 4 solutions i.e. (1, 4), (–1, 4), (1, –4) and (–1, –4)].
When x = ±2, y can take two values which is ±3. [Total of 4 solutions].
When x = ±3, y can take two values which is ±2. [Total 4 solutions].
When x = ±4, y can take two values which is ±1. [Total 4 solutions].
When x = ±5, y can take just one value which is 0. [Total of 2 solutions].
So, when we add up all the solutions, we get a total of 20 solutions.
Graphical Method:
We can also find out number integral points by plotting the graph.
So, we can find out number of integral points by counting all the integral points on four lines of the square. And answer can be found out orally by multiplying by 4 with whatever is the distance from the origin.
In this question also, we need to find the integral points, but only difference with previous question is that in this question we will have to count integral points on the four lines as well as in the region enclosed by four lines also.
We will learn this question how to count in such cases and after that such cases can be solved without plotting the graph.
Let us see how to count points.
In this question, we do not need to plot the graph if we have understood the previous question well.
We need to find integral points for less than 7, means that we have to count integral points for square made at 6, 5, 4, 3, 2, 1 and the origin.
We can just write the solutions for each of them and add them to get final solution.
Integral solutions for square made at 6: 6 × 4 = 24 integral solutions.
Integral solutions for square made at 5: 5 × 4 = 20 integral solutions
Integral solutions for square made at 4: 4 × 4 = 16 integral solutions.
Integral solutions for square made at 3: 3 × 4 = 12 integral solutions.
Integral solutions for square made at 2: 2 × 4 = 8 integral solutions.
Integral solutions for square made at 1: 1 × 4 = 4 integral solutions.
And don’t forget to count the origin (0, 0): 1 integral point.
On adding, we get total number of solutions as 105 integral solutions.
