Point A and B are in the first quadrant,point O is the origin. If the slope of OA is 1,slope of OB is 7 and OA=OB,Then slope of AB is: a. -1/5 b. -1/4 c. -1/3 d. -1/2

Options

A. `-1//5`
B. `-1//4`
C. `-1//3`
D. `-1//2`

Explanation

To solve the problem, we need to find the slope of line segment AB given the slopes of OA and OB and the condition that OA = OB. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the slopes
The slopes of the lines OA and OB are given as:
- Slope of OA (m_OA) = 1
- Slope of OB (m_OB) = 7

### Step 2: Determine angles from slopes
The slope of a line is given by the tangent of the angle it makes with the positive x-axis. Therefore:
- For OA:
\[ \tan(\alpha) = 1 \implies \alpha = 45^\circ \]
- For OB:
\[ \tan(\beta) = 7 \implies \beta = \tan^{-1}(7) \]

### Step 3: Express points A and B in Cartesian coordinates
Let OA = OB = r (since OA = OB). The coordinates of points A and B can be expressed as:
- Point A:
\[ A = (r \cos(45^\circ), r \sin(45^\circ)) = \left(\frac{r}{\sqrt{2}}, \frac{r}{\sqrt{2}}\right) \]
- Point B:
\[ B = (r \cos(\beta), r \sin(\beta)) \]

### Step 4: Find expressions for cos(β) and sin(β)
Using the relationship of tangent:
\[ \tan(\beta) = \frac{\sin(\beta)}{\cos(\beta)} = 7 \implies \sin(\beta) = 7 \cos(\beta) \]
Using the Pythagorean identity:
\[ \sin^2(\beta) + \cos^2(\beta) = 1 \]
Substituting \(\sin(\beta)\):
\[ (7 \cos(\beta))^2 + \cos^2(\beta) = 1 \implies 49 \cos^2(\beta) + \cos^2(\beta) = 1 \implies 50 \cos^2(\beta) = 1 \implies \cos^2(\beta) = \frac{1}{50} \]
Thus:
\[ \cos(\beta) = \frac{1}{5\sqrt{2}}, \quad \sin(\beta) = 7 \cos(\beta) = \frac{7}{5\sqrt{2}} \]

### Step 5: Substitute back to find coordinates of B
Now we can find the coordinates of point B:
\[ B = \left(r \cdot \frac{1}{5\sqrt{2}}, r \cdot \frac{7}{5\sqrt{2}}\right) = \left(\frac{r}{5\sqrt{2}}, \frac{7r}{5\sqrt{2}}\right) \]

### Step 6: Calculate the slope of line AB
The slope of line AB is given by:
\[ m_{AB} = \frac{y_B - y_A}{x_B - x_A} \]
Substituting the coordinates:
\[ m_{AB} = \frac{\frac{7r}{5\sqrt{2}} - \frac{r}{\sqrt{2}}}{\frac{r}{5\sqrt{2}} - \frac{r}{\sqrt{2}}} \]
Simplifying the numerator:
\[ = \frac{\frac{7r}{5\sqrt{2}} - \frac{5r}{5\sqrt{2}}}{\frac{r}{5\sqrt{2}} - \frac{5r}{5\sqrt{2}}} = \frac{\frac{2r}{5\sqrt{2}}}{-\frac{4r}{5\sqrt{2}}} \]
Cancelling \(r\) and \(5\sqrt{2}\):
\[ = \frac{2}{-4} = -\frac{1}{2} \]

### Final Answer
Thus, the slope of line segment AB is:
\[ \boxed{-\frac{1}{2}} \]