Answer :
Let's analyze each case to determine if the given points lie on a straight line.
### Case (a): Points [tex]\((5,0), (7,0),\)[/tex] and [tex]\((11,0)\)[/tex]
In this case, we need to check if the points [tex]\((5,0)\)[/tex], [tex]\((7,0)\)[/tex], and [tex]\((11,0)\)[/tex] are collinear.
1. Identify the coordinates:
- Point A: [tex]\((5, 0)\)[/tex]
- Point B: [tex]\((7, 0)\)[/tex]
- Point C: [tex]\((11, 0)\)[/tex]
2. Calculate the area of the triangle formed by these points:
The formula to check whether three points [tex]\((x_1, y_1)\)[/tex], [tex]\((x_2, y_2)\)[/tex], and [tex]\((x_3, y_3)\)[/tex] are collinear is by checking if the area of the triangle they form is zero.
The area of the triangle can be calculated using the determinant formula:
[tex]\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \][/tex]
Substituting the coordinates:
[tex]\[ \text{Area} = \frac{1}{2} \left| 5(0 - 0) + 7(0 - 0) + 11(0 - 0) \right| = \frac{1}{2} \times 0 = 0 \][/tex]
3. Conclusion:
Since the area is zero, the points [tex]\((5, 0)\)[/tex], [tex]\((7, 0)\)[/tex], and [tex]\((11, 0)\)[/tex] are collinear and lie on a straight line.
### Case (b): Points [tex]\((3,5), (3,6),\)[/tex] and [tex]\((3,15)\)[/tex]
In this case, we need to check if the points [tex]\((3,5)\)[/tex], [tex]\((3,6)\)[/tex], and [tex]\((3,15)\)[/tex] are collinear.
1. Identify the coordinates:
- Point P: [tex]\((3, 5)\)[/tex]
- Point Q: [tex]\((3, 6)\)[/tex]
- Point R: [tex]\((3, 15)\)[/tex]
2. Calculate the area of the triangle formed by these points:
Using the same area formula:
[tex]\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \][/tex]
Substituting the coordinates:
[tex]\[ \text{Area} = \frac{1}{2} \left| 3(6 - 15) + 3(15 - 5) + 3(5 - 6) \right| \][/tex]
[tex]\[ \text{Area} = \frac{1}{2} \left| 3(-9) + 3(10) + 3(-1) \right| \][/tex]
[tex]\[ \text{Area} = \frac{1}{2} \left| -27 + 30 - 3 \right| = \frac{1}{2} \left| 0 \right| = 0 \][/tex]
3. Conclusion:
Since the area is zero, the points [tex]\((3, 5)\)[/tex], [tex]\((3, 6)\)[/tex], and [tex]\((3, 15)\)[/tex] are collinear and lie on a straight line.
In both cases, the points given lie on a straight line.
### Case (a): Points [tex]\((5,0), (7,0),\)[/tex] and [tex]\((11,0)\)[/tex]
In this case, we need to check if the points [tex]\((5,0)\)[/tex], [tex]\((7,0)\)[/tex], and [tex]\((11,0)\)[/tex] are collinear.
1. Identify the coordinates:
- Point A: [tex]\((5, 0)\)[/tex]
- Point B: [tex]\((7, 0)\)[/tex]
- Point C: [tex]\((11, 0)\)[/tex]
2. Calculate the area of the triangle formed by these points:
The formula to check whether three points [tex]\((x_1, y_1)\)[/tex], [tex]\((x_2, y_2)\)[/tex], and [tex]\((x_3, y_3)\)[/tex] are collinear is by checking if the area of the triangle they form is zero.
The area of the triangle can be calculated using the determinant formula:
[tex]\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \][/tex]
Substituting the coordinates:
[tex]\[ \text{Area} = \frac{1}{2} \left| 5(0 - 0) + 7(0 - 0) + 11(0 - 0) \right| = \frac{1}{2} \times 0 = 0 \][/tex]
3. Conclusion:
Since the area is zero, the points [tex]\((5, 0)\)[/tex], [tex]\((7, 0)\)[/tex], and [tex]\((11, 0)\)[/tex] are collinear and lie on a straight line.
### Case (b): Points [tex]\((3,5), (3,6),\)[/tex] and [tex]\((3,15)\)[/tex]
In this case, we need to check if the points [tex]\((3,5)\)[/tex], [tex]\((3,6)\)[/tex], and [tex]\((3,15)\)[/tex] are collinear.
1. Identify the coordinates:
- Point P: [tex]\((3, 5)\)[/tex]
- Point Q: [tex]\((3, 6)\)[/tex]
- Point R: [tex]\((3, 15)\)[/tex]
2. Calculate the area of the triangle formed by these points:
Using the same area formula:
[tex]\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \][/tex]
Substituting the coordinates:
[tex]\[ \text{Area} = \frac{1}{2} \left| 3(6 - 15) + 3(15 - 5) + 3(5 - 6) \right| \][/tex]
[tex]\[ \text{Area} = \frac{1}{2} \left| 3(-9) + 3(10) + 3(-1) \right| \][/tex]
[tex]\[ \text{Area} = \frac{1}{2} \left| -27 + 30 - 3 \right| = \frac{1}{2} \left| 0 \right| = 0 \][/tex]
3. Conclusion:
Since the area is zero, the points [tex]\((3, 5)\)[/tex], [tex]\((3, 6)\)[/tex], and [tex]\((3, 15)\)[/tex] are collinear and lie on a straight line.
In both cases, the points given lie on a straight line.