Answer :
To find the equations of the lines [tex]\( l_1 \)[/tex] and [tex]\( l_2 \)[/tex] that pass through the point [tex]\( P(-4,0) \)[/tex] and are each 3 units away from the point [tex]\( A(2,-3) \)[/tex], we'll follow these steps:
1. Identify the conditions:
- The lines [tex]\( l_1 \)[/tex] and [tex]\( l_2 \)[/tex] pass through the point [tex]\( P(-4,0) \)[/tex].
- The distance from the point [tex]\( A(2,-3) \)[/tex] to each line must be 3 units.
2. Equation of a Line:
- The general equation of a line passing through the point [tex]\( P(-4, 0) \)[/tex] can be given by:
[tex]\[
y - 0 = m(x + 4)
\][/tex]
Simplifying, we get:
[tex]\[
y = m(x + 4)
\][/tex]
or rewritten in the standard form (for calculations involving distance):
[tex]\[
mx - y + 4m = 0
\][/tex]
3. Distance from a Point to a Line:
- The distance [tex]\( d \)[/tex] from a point [tex]\((x_1, y_1)\)[/tex] to a line given by [tex]\( ax + by + c = 0 \)[/tex] is calculated as:
[tex]\[
d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}
\][/tex]
- For point [tex]\( A(2, -3) \)[/tex] and each line, the formula becomes:
[tex]\[
\frac{|m \cdot 2 - (-3) + 4m|}{\sqrt{m^2 + (-1)^2}} = 3
\][/tex]
4. Solving for [tex]\( m \)[/tex]:
- The expression [tex]\( |2m + 3 + 4m| = |6m + 3| \)[/tex].
- So, we must solve:
[tex]\[
\frac{|6m + 3|}{\sqrt{m^2 + 1}} = 3
\][/tex]
- Solving gives two possible values for [tex]\( m \)[/tex]. This arises because the distance can be positive or negative depending on the line orientation, i.e., there are two lines perpendicular to vector direction.
5. Line Equations:
- Based on the solution of [tex]\( m \)[/tex], you'll find:
- For the first line [tex]\( l_1 \)[/tex]: Solve to find the appropriate [tex]\( m_1 \)[/tex].
- Construct the line equation for [tex]\( l_1 \)[/tex].
- Similarly, for the second line [tex]\( l_2 \)[/tex]: Solve to find [tex]\( m_2 \)[/tex].
- Construct [tex]\( l_2 \)[/tex] using its slope.
6. Conclusion:
- The equations derived represent lines [tex]\( l_1 \)[/tex] and [tex]\( l_2 \)[/tex] with the right conditions.
These equations capture the solution to the configuration described: lines that pass through a specific point and maintain a consistent distance from another point.
1. Identify the conditions:
- The lines [tex]\( l_1 \)[/tex] and [tex]\( l_2 \)[/tex] pass through the point [tex]\( P(-4,0) \)[/tex].
- The distance from the point [tex]\( A(2,-3) \)[/tex] to each line must be 3 units.
2. Equation of a Line:
- The general equation of a line passing through the point [tex]\( P(-4, 0) \)[/tex] can be given by:
[tex]\[
y - 0 = m(x + 4)
\][/tex]
Simplifying, we get:
[tex]\[
y = m(x + 4)
\][/tex]
or rewritten in the standard form (for calculations involving distance):
[tex]\[
mx - y + 4m = 0
\][/tex]
3. Distance from a Point to a Line:
- The distance [tex]\( d \)[/tex] from a point [tex]\((x_1, y_1)\)[/tex] to a line given by [tex]\( ax + by + c = 0 \)[/tex] is calculated as:
[tex]\[
d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}
\][/tex]
- For point [tex]\( A(2, -3) \)[/tex] and each line, the formula becomes:
[tex]\[
\frac{|m \cdot 2 - (-3) + 4m|}{\sqrt{m^2 + (-1)^2}} = 3
\][/tex]
4. Solving for [tex]\( m \)[/tex]:
- The expression [tex]\( |2m + 3 + 4m| = |6m + 3| \)[/tex].
- So, we must solve:
[tex]\[
\frac{|6m + 3|}{\sqrt{m^2 + 1}} = 3
\][/tex]
- Solving gives two possible values for [tex]\( m \)[/tex]. This arises because the distance can be positive or negative depending on the line orientation, i.e., there are two lines perpendicular to vector direction.
5. Line Equations:
- Based on the solution of [tex]\( m \)[/tex], you'll find:
- For the first line [tex]\( l_1 \)[/tex]: Solve to find the appropriate [tex]\( m_1 \)[/tex].
- Construct the line equation for [tex]\( l_1 \)[/tex].
- Similarly, for the second line [tex]\( l_2 \)[/tex]: Solve to find [tex]\( m_2 \)[/tex].
- Construct [tex]\( l_2 \)[/tex] using its slope.
6. Conclusion:
- The equations derived represent lines [tex]\( l_1 \)[/tex] and [tex]\( l_2 \)[/tex] with the right conditions.
These equations capture the solution to the configuration described: lines that pass through a specific point and maintain a consistent distance from another point.
