The sides of a rhombus ABCD are parallel to the lines `x−y+2=0` and `7x−y+3=0`, If the diagonals of the rhombus intersect at P(1,2) and the vertex A (different from the origin) is on the y-axis then the ordinate of A is:

Options

A. 2
B. `5/2`
C. `7/4`
D. `7/2`

Answer

To solve the problem, we need to find the ordinate of vertex A of the rhombus ABCD, given that the diagonals intersect at P(1, 2) and that vertex A lies on the y-axis. The sides of the rhombus are parallel to the lines given by the equations \(x - y + 2 = 0\) and \(7x - y + 3 = 0\).

### Step 1: Determine the slopes of the given lines
The first line can be rewritten in slope-intercept form:
\[ x - y + 2 = 0 \implies y = x + 2 \]
Thus, the slope \(m_1\) of the first line is \(1\).

The second line can also be rewritten:
\[ 7x - y + 3 = 0 \implies y = 7x + 3 \]
Thus, the slope \(m_2\) of the second line is \(7\).

### Step 2: Identify the slopes of the sides of the rhombus
Since the sides of the rhombus are parallel to these lines, the slopes of the sides of the rhombus will also be \(1\) and \(7\).

### Step 3: Find the coordinates of vertex A
Since vertex A lies on the y-axis, its coordinates can be represented as \(A(0, y_A)\). The diagonals of the rhombus bisect each other at point P(1, 2).

### Step 4: Use the midpoint formula
The midpoint of the diagonal connecting vertex A and the opposite vertex C must equal the coordinates of point P. Let the coordinates of vertex C be \(C(x_C, y_C)\). According to the midpoint formula:
\[ \left(\frac{0 + x_C}{2}, \frac{y_A + y_C}{2}\right) = (1, 2) \]
From this, we can set up two equations:
1. \(\frac{0 + x_C}{2} = 1\)
2. \(\frac{y_A + y_C}{2} = 2\)

### Step 5: Solve for \(x_C\)
From the first equation:
\[ x_C = 2 \]

### Step 6: Solve for \(y_C\)
From the second equation:
\[ y_A + y_C = 4 \implies y_C = 4 - y_A \]

### Step 7: Determine the slope condition
Since the sides of the rhombus are parallel to the lines with slopes \(1\) and \(7\), the slope of line AC (which connects A and C) can be calculated as follows:
\[ \text{slope of AC} = \frac{y_C - y_A}{x_C - 0} = \frac{(4 - y_A) - y_A}{2} = \frac{4 - 2y_A}{2} = 2 - y_A \]
This slope must be equal to either \(1\) or \(7\).

### Step 8: Set up equations for slopes
1. For slope \(1\):
\[ 2 - y_A = 1 \implies y_A = 1 \]

2. For slope \(7\):
\[ 2 - y_A = 7 \implies y_A = -5 \]

### Step 9: Determine the valid ordinate
Since vertex A is different from the origin and must lie on the y-axis, the ordinate of A cannot be negative (as it would place A below the x-axis). Therefore, the valid solution is:
\[ y_A = 1 \]

### Final Answer
The ordinate of vertex A is \(1\).

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