krit.club logo

Straight Lines - Distance of a Point From a Line

Grade 11CBSE

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

🔑Concepts

•

The distance from a point P(x1,y1)P(x_1, y_1) to a line LL is the length of the perpendicular segment dropped from the point to the line.

Perpendicular distance from a point to a straight line
•

Distance of the origin (0,0)(0, 0) from a line is the perpendicular distance calculated by substituting x1=0x_1=0 and y1=0y_1=0 into the distance formula.

Perpendicular distance from the origin to a line
•

Parallel lines have the same slope. The distance between two parallel lines is constant and is measured along a perpendicular transversing both lines.

•

To find the distance between parallel lines, we ensure both equations are in the form Ax+By+C=0Ax + By + C = 0 with identical AA and BB coefficients.

📐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=∣C1−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=∣c1−c2∣1+m2d = \frac{|c_1 - c_2|}{\sqrt{1 + m^2}}

💡Examples

Problem 1:

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

Solution:

  1. Identify the values from the point and the line equation: x1=3x_1 = 3, y1=−5y_1 = -5, A=3A = 3, B=−4B = -4, and C=−26C = -26.
  2. Substitute these values into the distance formula: d=∣3(3)+(−4)(−5)+(−26)∣32+(−4)2d = \frac{|3(3) + (-4)(-5) + (-26)|}{\sqrt{3^2 + (-4)^2}}
  3. Simplify the numerator: ∣9+20−26∣=∣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=0.6 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+8y−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=−34C_1 = -34, and C2=31C_2 = 31.
  2. Use the formula for distance between parallel lines: d=∣C1−C2∣A2+B2d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}
  3. Substitute the values: d=∣−34−31∣152+82d = \frac{|-34 - 31|}{\sqrt{15^2 + 8^2}}
  4. Simplify the numerator: ∣−65∣=65|-65| = 65
  5. Simplify the denominator: 225+64=289=17\sqrt{225 + 64} = \sqrt{289} = 17
  6. Calculate the final result: d=6517 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.

Problem 3:

Find the distance of the point (2,3)(2, 3) from the line 12x−5y+7=012x - 5y + 7 = 0.

Geometric representation of point (2,3) and the line 12x-5y+7=0

Solution:

  1. Identify coordinates and coefficients: x1=2,y1=3,A=12,B=−5,C=7x_1 = 2, y_1 = 3, A = 12, B = -5, C = 7.
  2. Apply the formula: d=∣Ax1+By1+C∣A2+B2d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}.
  3. Substitute values: d=∣12(2)+(−5)(3)+7∣122+(−5)2d = \frac{|12(2) + (-5)(3) + 7|}{\sqrt{12^2 + (-5)^2}}.
  4. Calculate: d=∣24−15+7∣144+25=16169=1613d = \frac{|24 - 15 + 7|}{\sqrt{144 + 25}} = \frac{16}{\sqrt{169}} = \frac{16}{13}. Distance is 1613\frac{16}{13} units.

Explanation:

We substitute the point (2,3)(2, 3) into the general equation of the line and divide by the magnitude of the normal vector (12,−5)(12, -5).

Problem 4:

Determine the distance between the parallel lines y=2x+1y = 2x + 1 and y=2x−9y = 2x - 9.

Two parallel lines with a perpendicular distance segment d

Solution:

  1. The lines are in slope-intercept form y=mx+cy = mx + c where m=2,c1=1,c2=−9m = 2, c_1 = 1, c_2 = -9.
  2. Use the formula: d=∣c1−c2∣1+m2d = \frac{|c_1 - c_2|}{\sqrt{1 + m^2}}.
  3. Substitute: d=∣1−(−9)∣1+22d = \frac{|1 - (-9)|}{\sqrt{1 + 2^2}}.
  4. Calculate: d=105=1055=25d = \frac{10}{\sqrt{5}} = \frac{10\sqrt{5}}{5} = 2\sqrt{5}. Distance is 252\sqrt{5} units.

Explanation:

Since the slopes are identical, we find the vertical gap between the y-intercepts and adjust for the tilt of the lines.