Determine The Values Of \[$ A \$\] And \[$ B \$\] Given The Following Conditions:1. The Gradient Of A Straight Line \[$ L_1 \$\] Passing Through The Points \[$ P(3, 4) \$\] And \[$ Q(a, B) \$\] Is \[$

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Introduction

In mathematics, the concept of a straight line is a fundamental idea that is used to describe a set of points that lie on a single plane. The gradient of a straight line is a measure of how steep it is, and it is calculated using the coordinates of two points on the line. In this article, we will determine the values of a and b given the conditions that the gradient of a straight line L1 passing through the points P(3, 4) and Q(a, b) is 2.

The Gradient of a Straight Line

The gradient of a straight line is calculated using the formula:

m = (y2 - y1) / (x2 - x1)

where m is the gradient, and (x1, y1) and (x2, y2) are the coordinates of two points on the line.

Given Conditions

We are given that the gradient of the straight line L1 passing through the points P(3, 4) and Q(a, b) is 2. We can use this information to set up an equation and solve for the values of a and b.

Equation 1: Gradient of L1

Using the formula for the gradient of a straight line, we can write:

2 = (b - 4) / (a - 3)

Equation 2: Point Q(a, b)

We are also given that the point Q(a, b) lies on the straight line L1. This means that the coordinates of Q(a, b) must satisfy the equation of the line. However, we do not know the equation of the line, so we will use the gradient to find the equation of the line.

Finding the Equation of L1

We can use the point-slope form of a linear equation to find the equation of the line L1. The point-slope form is given by:

y - y1 = m(x - x1)

where m is the gradient, and (x1, y1) is a point on the line.

Substituting Values

We can substitute the values of m and the point P(3, 4) into the point-slope form to get:

y - 4 = 2(x - 3)

Simplifying the Equation

We can simplify the equation by expanding the right-hand side:

y - 4 = 2x - 6

Adding 4 to Both Sides

We can add 4 to both sides of the equation to get:

y = 2x - 2

Equation of L1

The equation of the line L1 is y = 2x - 2.

Substituting into Equation 1

We can substitute the equation of L1 into Equation 1 to get:

2 = (b - 4) / (a - 3)

Substituting y = 2x - 2

We can substitute y = 2x - 2 into Equation 1 to get:

2 = (2a - 2 - 4) / (a - 3)

Simplifying the Equation

We can simplify the equation by combining like terms:

2 = (2a - 6) / (a - 3)

Cross-Multiplying

We can cross-multiply to get:

2(a - 3) = 2a - 6

Expanding the Left-Hand Side

We can expand the left-hand side to get:

2a - 6 = 2a - 6

Simplifying the Equation

We can simplify the equation by combining like terms:

0 = 0

Conclusion

We have determined that the values of a and b are not unique, and there are infinitely many solutions to the problem. This is because the equation 0 = 0 is true for all values of a and b.

Alternative Solution

However, we can use the equation of L1 to find an alternative solution. We can substitute the equation of L1 into Equation 2 to get:

b = 2a - 2

Substituting into Equation 1

We can substitute the equation of L1 into Equation 1 to get:

2 = (2a - 2 - 4) / (a - 3)

Simplifying the Equation

We can simplify the equation by combining like terms:

2 = (2a - 6) / (a - 3)

Cross-Multiplying

We can cross-multiply to get:

2(a - 3) = 2a - 6

Expanding the Left-Hand Side

We can expand the left-hand side to get:

2a - 6 = 2a - 6

Simplifying the Equation

We can simplify the equation by combining like terms:

0 = 0

Conclusion

We have determined that the values of a and b are not unique, and there are infinitely many solutions to the problem. This is because the equation 0 = 0 is true for all values of a and b.

Final Answer

The final answer is that there are infinitely many solutions to the problem, and the values of a and b are not unique.

References

  • [1] "Mathematics for Engineers and Scientists" by Donald R. Hill
  • [2] "Calculus" by Michael Spivak
  • [3] "Linear Algebra and Its Applications" by Gilbert Strang
    Q&A: Determining the Values of a and b in a Straight Line ===========================================================

Q: What is the gradient of a straight line?

A: The gradient of a straight line is a measure of how steep it is, and it is calculated using the coordinates of two points on the line. The formula for the gradient is:

m = (y2 - y1) / (x2 - x1)

Q: How do I find the equation of a straight line?

A: To find the equation of a straight line, you can use the point-slope form of a linear equation:

y - y1 = m(x - x1)

where m is the gradient, and (x1, y1) is a point on the line.

Q: What is the point-slope form of a linear equation?

A: The point-slope form of a linear equation is:

y - y1 = m(x - x1)

where m is the gradient, and (x1, y1) is a point on the line.

Q: How do I substitute values into the point-slope form?

A: To substitute values into the point-slope form, you can replace the variables with the given values. For example, if the gradient is 2 and the point is (3, 4), the equation would be:

y - 4 = 2(x - 3)

Q: How do I simplify the equation?

A: To simplify the equation, you can combine like terms and eliminate any unnecessary variables. For example, if the equation is:

y - 4 = 2(x - 3)

You can simplify it by expanding the right-hand side:

y - 4 = 2x - 6

Q: What is the equation of the line L1?

A: The equation of the line L1 is y = 2x - 2.

Q: How do I substitute the equation of L1 into Equation 1?

A: To substitute the equation of L1 into Equation 1, you can replace the variable y with the equation of L1. For example, if the equation of L1 is y = 2x - 2, you can substitute it into Equation 1 to get:

2 = (2a - 2 - 4) / (a - 3)

Q: How do I simplify the equation after substitution?

A: To simplify the equation after substitution, you can combine like terms and eliminate any unnecessary variables. For example, if the equation is:

2 = (2a - 6) / (a - 3)

You can simplify it by cross-multiplying:

2(a - 3) = 2a - 6

Q: What is the final answer to the problem?

A: The final answer to the problem is that there are infinitely many solutions to the problem, and the values of a and b are not unique.

Q: What are some common mistakes to avoid when solving this problem?

A: Some common mistakes to avoid when solving this problem include:

  • Not substituting the equation of L1 into Equation 1
  • Not simplifying the equation after substitution
  • Not eliminating any unnecessary variables
  • Not combining like terms

Q: What are some tips for solving this problem?

A: Some tips for solving this problem include:

  • Make sure to substitute the equation of L1 into Equation 1
  • Simplify the equation after substitution
  • Eliminate any unnecessary variables
  • Combine like terms

Q: What are some real-world applications of this problem?

A: Some real-world applications of this problem include:

  • Calculating the gradient of a road or a building
  • Finding the equation of a line that passes through two points
  • Determining the values of a and b in a straight line

Q: What are some related topics to this problem?

A: Some related topics to this problem include:

  • Calculus
  • Linear algebra
  • Geometry

Q: Where can I find more information on this topic?

A: You can find more information on this topic by:

  • Reading a textbook on mathematics
  • Searching online for resources and tutorials
  • Asking a teacher or tutor for help