To find the equations of the two straight lines drawn through the point (0, 1) such that the perpendicular dropped from the point (2, 2) to these lines is of unit length, we can follow these steps:
### Step 1: Identify the given points and the general equation of the line
We are given:
- Point \( P(0, 1) \) through which the lines pass.
- Point \( Q(2, 2) \) from which the perpendicular is dropped.
Let the equation of the line through point \( P \) be of the form:
\[
y = mx + c
\]
where \( m \) is the slope and \( c \) is the y-intercept.
### Step 2: Substitute the point (0, 1) into the line equation
Since the line passes through the point \( P(0, 1) \), we can substitute \( x = 0 \) and \( y = 1 \) into the line equation:
\[
1 = m(0) + c \implies c = 1
\]
Thus, the equation of the line simplifies to:
\[
y = mx + 1
\]
### Step 3: Use the formula for the length of the perpendicular from a point to a line
The length \( L \) of the perpendicular from a point \( (x_1, y_1) \) to the line \( Ax + By + C = 0 \) is given by:
\[
L = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}
\]
For our line \( y = mx + 1 \), we can rewrite it in the standard form:
\[
mx - y + 1 = 0
\]
Here, \( A = m \), \( B = -1 \), and \( C = 1 \).
Substituting \( (x_1, y_1) = (2, 2) \) into the formula, we have:
\[
L = \frac{|m(2) - (2) + 1|}{\sqrt{m^2 + (-1)^2}} = \frac{|2m - 2 + 1|}{\sqrt{m^2 + 1}} = \frac{|2m - 1|}{\sqrt{m^2 + 1}}
\]
### Step 4: Set the length of the perpendicular equal to 1
We know that the length of the perpendicular is 1 unit:
\[
\frac{|2m - 1|}{\sqrt{m^2 + 1}} = 1
\]
### Step 5: Solve the equation
Removing the absolute value, we have two cases:
1. \( 2m - 1 = \sqrt{m^2 + 1} \)
2. \( 2m - 1 = -\sqrt{m^2 + 1} \)
#### Case 1: \( 2m - 1 = \sqrt{m^2 + 1} \)
Squaring both sides:
\[
(2m - 1)^2 = m^2 + 1
\]
Expanding and simplifying:
\[
4m^2 - 4m + 1 = m^2 + 1 \implies 3m^2 - 4m = 0 \implies m(3m - 4) = 0
\]
Thus, \( m = 0 \) or \( m = \frac{4}{3} \).
#### Case 2: \( 2m - 1 = -\sqrt{m^2 + 1} \)
Squaring both sides:
\[
(2m - 1)^2 = m^2 + 1
\]
This leads to the same equation as Case 1, yielding the same values for \( m \).
### Step 6: Write the equations of the lines
1. For \( m = 0 \):
\[
y = 0 \cdot x + 1 \implies y = 1
\]
2. For \( m = \frac{4}{3} \):
\[
y = \frac{4}{3}x + 1
\]
### Final Answer
The equations of the two straight lines are:
1. \( y = 1 \)
2. \( y = \frac{4}{3}x + 1 \)
