https://file.aiquickdraw.com/mathbot/file/1726659847054w0gi90mg.jpg

Let's break down the problem shown in the image.

### Problem 46:

We are given two lines:
- Line ①: \( y = \frac{x}{3} \)
- Line ②: \( y = \frac{8}{x} \quad (x > 0) \)

There are two points:
- Point \( A(-6, 1) \)
- Point \( B(-1, -5) \)

The quadrilateral ABCD must be a parallelogram, and points \( C \) and \( D \) lie on lines ① and ②, respectively.

We are asked:
1. **(1)** If we set the x-coordinate of point \( C \), express the coordinates of point \( D \) using it.
2. **(2)** Find the exact coordinates of points \( C \) and \( D \).

#### Step-by-step Approach:

##### For Question 1:
Let the x-coordinate of point \( C \) be \( x_C \). Since \( C \) lies on line ①, its y-coordinate can be written as:
\[ y_C = \frac{x_C}{3} \]
So, the coordinates of point \( C \) are:
\[ C(x_C, \frac{x_C}{3}) \]

Now, since ABCD is a parallelogram, vector \( \overrightarrow{AB} \) must be equal to vector \( \overrightarrow{CD} \). 
We know:
\[ \overrightarrow{AB} = B - A = (-1, -5) - (-6, 1) = (5, -6) \]
Thus, we have:
\[ \overrightarrow{CD} = (5, -6) \]

Using \( C(x_C, \frac{x_C}{3}) \), we find the coordinates of \( D \) by adding vector \( \overrightarrow{CD} \):
\[ D = C + \overrightarrow{CD} = \left(x_C + 5, \frac{x_C}{3} - 6\right) \]

Since point \( D \) lies on line ② (\( y = \frac{8}{x} \)), we can substitute the y-coordinate of \( D \) into the equation for line ②:
\[
\frac{x_C}{3} - 6 = \frac{8}{x_C + 5}
\]

##### For Question 2:
Solving the equation above will give us the exact value of \( x_C \), which will help us find both points \( C \) and \( D \).

Let's solve this system of equations next.

#### Solving for \( x_C \):
We have the equation:
\[
\frac{x_C}{3} - 6 = \frac{8}{x_C + 5}
\]

Let me solve this step-by-step.

I'll now calculate this.The solutions for \( x_C \) are \( x_C = -6 \) and \( x_C = 19 \).

Since point \( C \) lies on line ① where \( x > 0 \), we discard \( x_C = -6 \) and choose \( x_C = 19 \).

### Coordinates of Point \( C \):
Using \( x_C = 19 \), the coordinates of point \( C \) are:
\[
C(19, \frac{19}{3}) = C(19, 6.33)
\]

### Coordinates of Point \( D \):
Now, using \( x_C = 19 \) in the expression for \( D \):
\[
D = \left( 19 + 5, \frac{19}{3} - 6 \right) = D(24, 0.33)
\]

Thus, the coordinates of points \( C \) and \( D \) are:
- \( C(19, 6.33) \)
- \( D(24, 0.33) \)

Would you like further details on this solution, or should I proceed with the next problem?

### Related Questions:
1. How would the solution change if the parallelogram was not aligned with the given axes?
2. Can you express the area of the parallelogram in terms of the coordinates of points \( C \) and \( D \)?
3. What if the line \( y = \frac{8}{x} \) had \( x < 0 \)? How would that affect the solution?
4. How can we determine the angles of this parallelogram from the points?
5. Can this problem be solved graphically or with a geometric software tool?

### Tip:
When solving geometry problems, sketching the problem and labeling all known points can provide clarity and help prevent calculation errors.

右の図のように、直線① y = x / 3, 双曲線② y = 8 / x (x > 0), と、2点A(-6, 1), B(-1, -5) がある。また、四角形ABCDが平行四辺形となるように、2点C, Dをそれぞれ①、②の上にとる。次の問いに答えなさい。(1) 点Cのx座標をtとおくとき、点Dの座標を用いて表しなさい。(2) 2点C, Dの座標をそれぞれ求めなさい。

Solve a coordinate geometry problem involving a parallelogram ABCD, where points C and D lie on a line and a hyperbola respectively. Learn how to express coordinates and use vectors to find the exact solution.

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