To find the equation of the line passing through the center and bisecting the chord \(7x + y - 1 = 0\) of the ellipse \(\frac{x^2}{1} + \frac{y^2}{7} = 1\), we can follow these steps:
### Step 1: Identify the center of the ellipse
The standard form of the ellipse given is \(\frac{x^2}{1} + \frac{y^2}{7} = 1\). The center of this ellipse is at the origin, which is \((0, 0)\).
**Hint:** The center of an ellipse in standard form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) is always at the point \((0, 0)\).
### Step 2: Find the midpoint of the chord
The chord is given by the equation \(7x + y - 1 = 0\). To find the midpoint of this chord, we can express \(y\) in terms of \(x\):
\[
y = 1 - 7x
\]
The midpoint of the chord can be found by substituting values of \(x\) and \(y\) that satisfy this equation. However, we will use the general property of chords in ellipses.
### Step 3: Use the chord midpoint formula
For the chord \(Ax + By + C = 0\), the midpoint \((H, K)\) can be determined by the formula:
\[
\frac{H}{a^2} = \frac{A}{B}, \quad \frac{K}{b^2} = \frac{B}{A}
\]
Here, \(A = 7\), \(B = 1\), \(C = -1\), \(a^2 = 1\), and \(b^2 = 7\).
### Step 4: Set up the equations
From the properties of the ellipse, we can set up the equations:
\[
\frac{H}{1} = \frac{7}{1} \quad \text{and} \quad \frac{K}{7} = \frac{1}{7}
\]
This simplifies to:
\[
H = 7K
\]
**Hint:** Remember that for a chord of an ellipse, the midpoint coordinates can be expressed in terms of the coefficients of the chord equation.
### Step 5: Solve for \(H\) and \(K\)
From the equations \(H = 7K\) and knowing that \(H\) and \(K\) are equal (from the properties of the ellipse), we can set \(H = K\). Therefore, we have:
\[
H = 7H \implies H(1 - 7) = 0 \implies H = 0
\]
Thus, \(K = 0\).
### Step 6: Write the equation of the line
Since we have \(H = 0\) and \(K = 0\), the line that passes through the center and bisects the chord is given by:
\[
x - y = 0 \quad \text{or} \quad x = y
\]
### Final Answer
The equation of the line passing through the center and bisecting the chord \(7x + y - 1 = 0\) of the ellipse \(\frac{x^2}{1} + \frac{y^2}{7} = 1\) is:
\[
x - y = 0
\]
