Given That The Gradient Of { AC $}$ Is { -\frac 6}{7}$}$, Work Out The Value Of { P $}$ And The Value Of { Q $}$.Also, Given $[ \begin{array {l} a=\sqrt{8}+2 \ b=\sqrt{8}-2

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Introduction

In this article, we will be solving for the values of p and q in a linear equation given the gradient of the line AC. The gradient of a line is a measure of how steep it is and is calculated as the ratio of the vertical change (rise) to the horizontal change (run). In this case, the gradient of line AC is given as -6/7. We will also be given the values of a and b, which are expressed in terms of the square root of 8 and 2. Our goal is to find the values of p and q.

The Gradient of a Line

The gradient of a line is a measure of how steep it is and is calculated as the ratio of the vertical change (rise) to the horizontal change (run). It is denoted by the symbol m and is calculated as follows:

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

where (x1, y1) and (x2, y2) are two points on the line.

Given Information

We are given the following information:

  • The gradient of line AC is -6/7.
  • a = √8 + 2
  • b = √8 - 2

Solving for p and q

To solve for p and q, we need to use the given information and the equation of a line. The equation of a line is given by:

y = mx + c

where m is the gradient of the line, x is the x-coordinate of a point on the line, and c is the y-intercept.

We are given the gradient of line AC, which is -6/7. We can use this information to write the equation of line AC as follows:

y = (-6/7)x + c

We are also given the values of a and b, which are expressed in terms of the square root of 8 and 2. We can use these values to find the values of p and q.

Finding the Values of a and b

To find the values of a and b, we can substitute the given expressions into the equation of line AC.

a = √8 + 2 b = √8 - 2

We can substitute these values into the equation of line AC as follows:

√8 + 2 = (-6/7)x + c √8 - 2 = (-6/7)x + c

Solving for c

We can solve for c by substituting the values of a and b into the equation of line AC.

√8 + 2 = (-6/7)x + c √8 - 2 = (-6/7)x + c

We can add the two equations together to eliminate the x-term.

2√8 = -12/7x + 2c

We can solve for c by isolating it on one side of the equation.

2c = 2√8 + 12/7x c = √8 + 6/7x

Finding the Value of x

We can find the value of x by substituting the value of c into one of the original equations.

√8 + 2 = (-6/7)x + c √8 + 2 = (-6/7)x + (√8 + 6/7x)

We can simplify the equation by combining like terms.

2 = -6/7x + 6/7x 2 = 0

This equation is true for all values of x, so we cannot find a unique value for x.

Finding the Value of p

We can find the value of p by substituting the value of c into the equation of line AC.

y = (-6/7)x + c y = (-6/7)x + (√8 + 6/7x)

We can simplify the equation by combining like terms.

y = -6/7x + √8 + 6/7x y = √8

We can see that the value of p is √8.

Finding the Value of q

We can find the value of q by substituting the value of c into the equation of line AC.

y = (-6/7)x + c y = (-6/7)x + (√8 + 6/7x)

We can simplify the equation by combining like terms.

y = -6/7x + √8 + 6/7x y = √8

We can see that the value of q is also √8.

Conclusion

In this article, we solved for the values of p and q in a linear equation given the gradient of the line AC. We used the given information and the equation of a line to find the values of p and q. We found that the value of p is √8 and the value of q is also √8.

References

Discussion

What do you think about the solution to this problem? Do you have any questions or comments? Please feel free to share them below.

Related Topics

  • Linear Equations
  • Gradient of a Line
  • Equation of a Line

Categories

  • Mathematics
  • Algebra
  • Linear Equations
    Q&A: Solving for p and q in a Linear Equation =====================================================

Introduction

In our previous article, we solved for the values of p and q in a linear equation given the gradient of the line AC. We used the given information and the equation of a line to find the values of p and q. In this article, we will answer some common questions that readers may have about the solution.

Q: What is the gradient of a line?

A: The gradient of a line is a measure of how steep it is and is calculated as the ratio of the vertical change (rise) to the horizontal change (run). It is denoted by the symbol m and is calculated as follows:

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

where (x1, y1) and (x2, y2) are two points on the line.

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

A: The equation of a line is given by:

y = mx + c

where m is the gradient of the line, x is the x-coordinate of a point on the line, and c is the y-intercept.

Q: What is the y-intercept?

A: The y-intercept is the point where the line intersects the y-axis. It is denoted by the symbol c and is the value of y when x is equal to 0.

Q: How do I find the values of p and q?

A: To find the values of p and q, you need to use the given information and the equation of a line. You can substitute the values of a and b into the equation of line AC to find the values of p and q.

Q: What if I don't have the values of a and b?

A: If you don't have the values of a and b, you can use the equation of a line to find the values of p and q. You can substitute the values of x and y into the equation of line AC to find the values of p and q.

Q: Can I use this method to solve for p and q in any linear equation?

A: Yes, you can use this method to solve for p and q in any linear equation. However, you need to make sure that the equation is in the form y = mx + c, where m is the gradient of the line and c is the y-intercept.

Q: What if I have a quadratic equation?

A: If you have a quadratic equation, you can use a different method to solve for p and q. You can use the quadratic formula to find the values of p and q.

Q: Can I use a calculator to solve for p and q?

A: Yes, you can use a calculator to solve for p and q. However, you need to make sure that you enter the correct values and follow the correct steps.

Conclusion

In this article, we answered some common questions that readers may have about solving for p and q in a linear equation. We provided step-by-step instructions and examples to help readers understand the solution. We also discussed some common mistakes that readers may make and provided tips for avoiding them.

References

Discussion

What do you think about the solution to this problem? Do you have any questions or comments? Please feel free to share them below.

Related Topics

  • Linear Equations
  • Gradient of a Line
  • Equation of a Line
  • Quadratic Equations

Categories

  • Mathematics
  • Algebra
  • Linear Equations