Distance of a line from a given point
Drawing upon the information from the sources and our conversation history, we can discuss the distance of a point from a given line, particularly when the line is expressed in its General Form.
The problem is to determine the distance of a point P, with coordinates (x₁, y₁), from a line l, given by the equation Ax + By + C = 0. This form, Ax + By + C = 0, is known as the general form of the equation of a line, and it is a versatile representation as it can describe any straight line [Conversation history].
The distance referred to in this context is the shortest distance from the point to the line. Geometrically, this shortest distance is the length of the perpendicular segment dropped from the point to the line. Let M be the point on the line such that PM is perpendicular to the line; the distance is the length of PM.
To find this distance, the sources outline a method involving the geometric properties of the line. One approach discussed is to consider the triangle formed by the given point P(x₁, y₁) and the points where the line intersects the x and y axes (the x-intercept and y-intercept). If the line Ax + By + C = 0 (where A ≠ 0 and B ≠ 0) is written in the intercept form (x/a + y/b = 1), the x-intercept is at Q(-C/A, 0) and the y-intercept is at R(0, -C/B). The line l is the line passing through Q and R.
The distance PM (the perpendicular distance from P to the line QR) can be found using the formula for the area of a triangle: Area of triangle PQR = (1/2) * base * height. Using QR as the base and PM as the height, we get Area = (1/2) * QR * PM. Therefore, the distance PM = (2 * Area of triangle PQR) / QR. The length of the base QR can be calculated using the distance formula between points Q(-C/A, 0) and R(0, -C/B). The area of the triangle PQR can be calculated using the coordinates of its vertices P(x₁, y₁), Q(-C/A, 0), and R(0, -C/B).
Following this process, the sources provide the resulting formula for the distance (d) of a straight line ax + by + c = 0 from a point (x₁, y₁):
d = |ax₁ + by₁ + c| / √(a² + b²)
This formula works directly with the coefficients (a, b, c) of the general form of the line and the coordinates (x₁, y₁) of the point. The numerator is the absolute value of the expression ax₁ + by₁ + c, obtained by substituting the point’s coordinates into the line’s equation. The denominator is the square root of the sum of the squares of the coefficients of x and y.
An example calculation is provided for the distance of the point (3, -5) from the line 3x - 4y + 12 = 0. Using the formula: a = 3, b = -4, c = 12 x₁ = 3, y₁ = -5 d = |(3)(3) + (-4)(-5) + 12| / √(3² + (-4)²) d = |9 + 20 + 12| / √(9 + 16) d = |41| / √25 d = 41 / 5 (Note: Source states “the numerator will be 3”, which appears to be a transcription error as the calculation shows the numerator is 41). d = 8.2.
This formula is a fundamental result derived from coordinate geometry. The concept of the distance of a point from a line is also useful in calculating the distance between two parallel lines.
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