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Solve Lesson 8 Page 25 SBT Math 10 – Kite>

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Topic

The uncrossed half-plane (excluding d) in each of Figures 5a, 5b, 5c is the domain of which of the equations?

Solution method – See details

  • Step 1: Determine the equation of the line dividing the plane into two parts of the form \(ax + by = c\)
  • Step 2: Take a point \(M\left( {{x_o};{y_o}} \right)\) in the solution domain of the inequality, replace the coordinates of the point M into \(ax + by\) and compare Compare with c to determine the required inequality

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Detailed explanation

+ Figure 8a): The line d is parallel to the axis Ox and passes through the point (0; 2), so d is y = 2 or 0.x + 1.y = 2.

Taking O(0; 0) has 0.0 + 1.0 = 0 < 2.

Looking at Figure 8a), we see that the point O(0; 0) does not belong to the domain of the solution of the inequality and does not include the line d, so the inequality to be found is: y > 2.

So the solution domain inequality shown in Figure 8a) is y > 2.

+ Figure 8b): The line d is parallel to the Oy axis and passes through the point (1; 0) so d is x = 1 or x + 0.y = 1.

Taking O(0; 0) has 1.0 + 0.0 = 0 < 1.

Looking at Figure 8b), we see that the point O(0; 0) belongs to the domain of the solution of the inequality and does not include the line d, so the inequality to be found is: x < 1.

So the solution domain inequality shown in Figure 8b) is x < 1.

+) Figure 8c): Let the equation of the line d have the form: \(y = ax + b\left( {a \ne 0} \right)\)

The line d is a straight line that passes through two points with coordinates (– 2; 0), so replacing the coordinates of this point in the equation d we get: \(0 = a.\left( { – 2} \right) + b \Leftrightarrow – 2a + b = 0\left( 1 \right)\)

The line d is a line that passes through two points with coordinates (0; – 1), so replacing the coordinates of this point into the equation d we get: \( – 1 = a.0 + b \Leftrightarrow b = – 1\)

Substituting b = 0 – 1 into (1) we get \( – 2a + \left( { – 1} \right) = 0 \Leftrightarrow a = – \frac{1}{2}\).

Derive the equation of the line d as \(y = – \frac{1}{2}x – 1\) or \(\frac{1}{2}x + y = – 1\)

Take O(0; 0) with \(\frac{1}{2}.0 + 0 = 0 > – 1\)

Looking at Figure 8c), we see that the point O(0; 0) in the half-plane is the solution domain of the inequality and excluding the line d, so the inequality to be found is: \(\frac{1}{2} x + y > – 1\)

So the solution domain inequality shown in Figure 8c) is \(\frac{1}{2}x + y > – 1\)

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