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Difficulty: HardCoordinate Geometry of Straight Lines

A straight line L1L_1 has the equation 3x−4y+5=03x - 4y + 5 = 0. A second line L2L_2 is parallel to L1L_1 and passes through the point (6,1)(6, 1). What is the perpendicular distance between lines L1L_1 and L2L_2?

Answer: 3.8

Answer

The perpendicular distance between lines L1L_1 and L2L_2 is 3.83.8 units.
The perpendicular distance between 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}}. Line L2L_2 is parallel to 3x−4y+5=03x - 4y + 5 = 0, so its equation is 3x−4y+C=03x - 4y + C = 0. Substituting (6,1)(6, 1) gives 3(6)−4(1)+C=03(6) - 4(1) + C = 0, leading to C=−14C = -14. Substituting C1=5C_1 = 5 and C2=−14C_2 = -14 into the distance formula gives d=∣5−(−14)∣32+(−4)2=195=3.8d = \frac{|5 - (-14)|}{\sqrt{3^2 + (-4)^2}} = \frac{19}{5} = 3.8.

Step-by-Step Solution

1
Determine the equation of line L2L_2
The equation of L2L_2 is 3x−4y−14=03x - 4y - 14 = 0
Lines parallel to 3x−4y+5=03x - 4y + 5 = 0 have the form 3x−4y+C=03x - 4y + C = 0. Substituting the point (6,1)(6, 1) gives 3(6)−4(1)+C=0  ⟹  C=−143(6) - 4(1) + C = 0 \implies C = -14.
2
Apply the parallel line distance formula
d=∣5−(−14)∣32+(−4)2=195d = \frac{|5 - (-14)|}{\sqrt{3^2 + (-4)^2}} = \frac{19}{5}
The perpendicular distance between parallel lines Ax+By+C1=0Ax + By + C_1 = 0 and Ax+By+C2=0Ax + By + C_2 = 0 is given by d=∣C1−C2∣A2+B2d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}.
3
Convert fraction to decimal form
3.83.8
Dividing 1919 by 55 yields 3.83.8.

Key Concept

Perpendicular Distance Between Parallel Lines
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Topics

Coordinate Geometry of Straight Lines

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