Find the direction in which a straight line must be drawn through the point so that its point of intersection with the line may be at a distance of 3 units from this point.
The line must be drawn parallel to the x-axis (making angle 0° with positive x-axis) or parallel to the y-axis (making angle 90° with positive x-axis). The equations of such lines are y = 2 and x = -1 respectively.
Explanation
This question is similar to Example 12 in the textbook context, where we find the intersection point and use distance formula. The key insight is that points at distance 3 from (-1, 2) lie on a circle. Finding where this circle intersects the line x + y = 4 gives us the required intersection points, from which we determine the direction of lines through (-1, 2).
Solution Steps
Step 1: Points at distance 3 from (-1, 2) lie on the circle (x + 1)² + (y - 2)² = 9.
Step 2: Find intersection of this circle with line x + y = 4. Substitute y = 4 - x into circle equation: (x + 1)² + (2 - x)² = 9
Step 3: Simplify: x² + 2x + 1 + 4 - 4x + x² = 9, giving 2x² - 2x - 4 = 0, or x² - x - 2 = 0
Step 4: Solve: (x - 2)(x + 1) = 0, so x = 2 or x = -1
Step 5: Corresponding points: (2, 2) and (-1, 5)
Step 6: Line through (-1, 2) and (2, 2): slope = 0, horizontal line y = 2 (angle 0°)
Step 7: Line through (-1, 2) and (-1, 5): slope undefined, vertical line x = -1 (angle 90°)