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Straight Lines - Distance of a Point From a Line

Grade 11CBSE

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

πŸ”‘Concepts

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The distance of a point from a line is defined as the length of the perpendicular segment drawn from the point to the line. Visually, if you have a line LL and a point PP not on the line, the shortest path to the line is the straight segment PMPM that intersects LL at a 90∘90^{\circ} angle at the foot of the perpendicular MM.

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To calculate distance, the line must be expressed in the general form Ax+By+C=0Ax + By + C = 0. In a coordinate plane, the coefficients AA and BB relate to the slope of the line, while CC determines its position relative to the origin.

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The distance is always considered a non-negative scalar quantity. Mathematically, this is handled by using the modulus (absolute value) symbol ∣...∣|...| in the numerator of the formula, ensuring that even if the coordinates produce a negative result, the final distance is positive.

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The distance from the origin (0,0)(0, 0) to a line is a simplification where the point coordinates are zero. Geometrically, this represents the length of the perpendicular segment from the center of the coordinate system (0,0)(0, 0) to the nearest point on the line Ax+By+C=0Ax + By + C = 0.

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Parallel lines are lines that share the same slope and never intersect. On a graph, they appear as two lines following the same direction with a constant gap between them. Their equations can be written as Ax+By+C1=0Ax + By + C_1 = 0 and Ax+By+C2=0Ax + By + C_2 = 0, where the xx and yy coefficients are identical.

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The distance between two parallel lines is the perpendicular distance from any point on one line to the other line. This distance remains constant regardless of which point is chosen on the first line, as the lines never converge or diverge.

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If the calculated distance of a point from a line is zero, it implies that the point (x1,y1)(x_1, y_1) satisfies the equation Ax1+By1+C=0Ax_1 + By_1 + C = 0, meaning the point lies directly on the line itself.

πŸ“Formulae

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

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

The distance dd between two parallel lines Ax+By+C1=0Ax + By + C_1 = 0 and Ax+By+C2=0Ax + By + C_2 = 0 is: d=∣C1βˆ’C2∣A2+B2d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}

The distance dd between two parallel lines in slope-intercept form y=mx+c1y = mx + c_1 and y=mx+c2y = mx + c_2 is: d=∣c1βˆ’c2∣1+m2d = \frac{|c_1 - c_2|}{\sqrt{1 + m^2}}

πŸ’‘Examples

Problem 1:

Find the distance of the point (3,βˆ’5)(3, -5) from the line 3xβˆ’4yβˆ’26=03x - 4y - 26 = 0.

Solution:

  1. Identify the values from the point and the line equation: x1=3x_1 = 3, y1=βˆ’5y_1 = -5, A=3A = 3, B=βˆ’4B = -4, and C=βˆ’26C = -26.
  2. Substitute these values into the distance formula: d=∣3(3)+(βˆ’4)(βˆ’5)+(βˆ’26)∣32+(βˆ’4)2d = \frac{|3(3) + (-4)(-5) + (-26)|}{\sqrt{3^2 + (-4)^2}}
  3. Simplify the numerator: ∣9+20βˆ’26∣=∣3∣=3|9 + 20 - 26| = |3| = 3
  4. Simplify the denominator: 9+16=25=5\sqrt{9 + 16} = \sqrt{25} = 5
  5. Calculate the final distance: d=35=0.6Β unitsd = \frac{3}{5} = 0.6 \text{ units}

Explanation:

We use the standard distance formula for a point to a line. The absolute value ensures the distance is positive, and the denominator represents the magnitude of the normal vector to the line.

Problem 2:

Find the distance between the parallel lines 15x+8yβˆ’34=015x + 8y - 34 = 0 and 15x+8y+31=015x + 8y + 31 = 0.

Solution:

  1. Identify the coefficients: A=15A = 15, B=8B = 8, C1=βˆ’34C_1 = -34, and C2=31C_2 = 31.
  2. Use the formula for distance between parallel lines: d=∣C1βˆ’C2∣A2+B2d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}
  3. Substitute the values: d=βˆ£βˆ’34βˆ’31∣152+82d = \frac{|-34 - 31|}{\sqrt{15^2 + 8^2}}
  4. Simplify the numerator: βˆ£βˆ’65∣=65|-65| = 65
  5. Simplify the denominator: 225+64=289=17\sqrt{225 + 64} = \sqrt{289} = 17
  6. Calculate the final result: d=6517Β unitsd = \frac{65}{17} \text{ units}

Explanation:

Since the xx and yy coefficients are the same for both lines, they are parallel. The distance between them is the difference in their constants divided by the square root of the sum of the squares of the coefficients.