Finding a Point on the Line x - y + 1 = 0
We are given a line represented by the equation x - y + 1 = 0 and a point (1, 2). Our goal is to find a point on the line that is 2√2 units away from the given point.
Understanding the Problem
- Line Equation: The equation x - y + 1 = 0 defines a straight line in the x-y plane.
- Distance: We need to find a point on this line that is exactly 2√2 units away from the point (1, 2).
Solution Approach
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Find the slope of the line: Rearranging the equation, we get y = x + 1. This tells us that the slope of the line is 1.
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Find the perpendicular slope: The slope of a line perpendicular to the given line will be the negative reciprocal of its slope. Therefore, the perpendicular slope is -1.
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Find the equation of the perpendicular line: We know that this perpendicular line passes through the point (1, 2). Using the point-slope form (y - y1 = m(x - x1)), we get the equation:
y - 2 = -1(x - 1) y = -x + 3
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Find the intersection point: We now have two equations:
- x - y + 1 = 0 (original line)
- y = -x + 3 (perpendicular line)
Solving this system of equations, we can find the point of intersection. Substituting the second equation into the first equation, we get:
x - (-x + 3) + 1 = 0 2x - 2 = 0 x = 1
Substituting x = 1 into the equation y = -x + 3, we find y = 2.
Therefore, the intersection point is (1, 2). This is the same as the given point!
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Finding the desired points: Since the intersection point is the same as the given point, it means that the line is perpendicular to the line segment connecting the given point to any point on the line x - y + 1 = 0. To find the points on the line that are 2√2 units away from the given point, we need to move along the line x - y + 1 = 0 from the intersection point (1, 2) in both directions.
We can use the distance formula to find the x-coordinate of these points:
2√2 = √((x - 1)² + (y - 2)²)
Substituting y = x + 1 (from the equation of the original line) into the distance formula, we get:
2√2 = √((x - 1)² + (x - 1)²)
Solving for x, we find two solutions: x = 1 + √2 and x = 1 - √2.
Finally, we can find the corresponding y-coordinates using the equation y = x + 1:
- For x = 1 + √2, y = 2 + √2
- For x = 1 - √2, y = 2 - √2
Therefore, the two points on the line x - y + 1 = 0 that are 2√2 units away from the point (1, 2) are (1 + √2, 2 + √2) and (1 - √2, 2 - √2).
