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Coordinate Geometry - Distance of a Point from a Line

Grade 11ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Perpendicular Distance: The distance of a point P(x1,y1)P(x_1, y_1) from a line Ax+By+C=0Ax + By + C = 0 is defined as the shortest path from the point to the line. Visually, if you drop a perpendicular segment from point PP to the line, the length of this segment represents the distance. It forms a 90∘90^{\circ} angle at the point of intersection on the line.

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The General Equation Form: To find the distance, the line must be expressed in its general form Ax+By+C=0Ax + By + C = 0. In a coordinate plane, the coefficients AA and BB are related to the direction of the line, while CC dictates its position relative to the origin.

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Distance from the Origin: This is a special case where the point is the origin (0,0)(0, 0). Visually, this is the length of the normal (perpendicular) drawn from the intersection of the xx and yy axes to the line. If the line passes through (0,0)(0,0), the constant CC will be zero, making the distance zero.

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Distance Between Parallel Lines: Two lines are parallel if they have the same slope, represented as Ax+By+C1=0Ax + By + C_1 = 0 and Ax+By+C2=0Ax + By + C_2 = 0. Visually, these lines look like straight tracks that never meet. The distance between them is the constant perpendicular gap measured anywhere along the lines.

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Position of a Point: Substituting the coordinates of a point (x1,y1)(x_1, y_1) into the expression Ax+By+CAx + By + C allows us to determine if a point lies on the line (result is 00) or which side of the line it occupies. On a graph, points that produce the same sign lie on the same side of the line.

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Normal Form of a Line: A line can be described as xcos⁡α+ysin⁡α=px \cos \alpha + y \sin \alpha = p, where pp is the length of the perpendicular from the origin and α\alpha is the angle that this perpendicular makes with the positive xx-axis. This highlights the geometric relationship between the line's orientation and its distance from the center.

📐Formulae

Distance dd of point (x1,y1)(x_1, y_1) from line Ax+By+C=0Ax + By + C = 0: d=∣Ax1+By1+C∣A2+B2d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}

Distance dd of the origin (0,0)(0, 0) from line Ax+By+C=0Ax + By + C = 0: d=∣C∣A2+B2d = \frac{|C|}{\sqrt{A^2 + B^2}}

Distance dd between two parallel lines Ax+By+C1=0Ax + By + C_1 = 0 and Ax+By+C2=0Ax + By + C_2 = 0: d=∣C1−C2∣A2+B2d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}

Slope mm of the line Ax+By+C=0Ax + By + C = 0: m=−ABm = -\frac{A}{B}

💡Examples

Problem 1:

Find the perpendicular distance of the point P(3,−5)P(3, -5) from the line 3x−4y−26=03x - 4y - 26 = 0.

Solution:

Step 1: Identify the values from the point and the line equation. Here, x1=3x_1 = 3, y1=−5y_1 = -5, A=3A = 3, B=−4B = -4, and C=−26C = -26.

Step 2: Apply the distance formula: d=∣Ax1+By1+C∣A2+B2d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}

Step 3: Substitute the values: d=∣3(3)+(−4)(−5)+(−26)∣32+(−4)2d = \frac{|3(3) + (-4)(-5) + (-26)|}{\sqrt{3^2 + (-4)^2}} d=∣9+20−26∣9+16d = \frac{|9 + 20 - 26|}{\sqrt{9 + 16}} d=∣3∣25d = \frac{|3|}{\sqrt{25}} d=35=0.6d = \frac{3}{5} = 0.6 units.

Explanation:

We use the standard distance formula by plugging in the coordinates of the point into the line's equation in the numerator and dividing by the magnitude of the line's normal vector in the denominator.

Problem 2:

Find the distance between the parallel lines 5x+12y−20=05x + 12y - 20 = 0 and 5x+12y+19=05x + 12y + 19 = 0.

Solution:

Step 1: Identify the coefficients. Since the lines are parallel, A=5A = 5 and B=12B = 12 for both. The constants are C1=−20C_1 = -20 and C2=19C_2 = 19.

Step 2: Apply the formula for the distance between parallel lines: d=∣C1−C2∣A2+B2d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}

Step 3: Substitute the values: d=∣−20−19∣52+122d = \frac{|-20 - 19|}{\sqrt{5^2 + 12^2}} d=∣−39∣25+144d = \frac{|-39|}{\sqrt{25 + 144}} d=39169d = \frac{39}{\sqrt{169}} d=3913=3d = \frac{39}{13} = 3 units.

Explanation:

To find the distance between parallel lines, we calculate the absolute difference between their constant terms and divide by the square root of the sum of the squares of the xx and yy coefficients.