Class 11 Mathematics | Written by Neeraj Anand
Published by ANAND TECHNICAL PUBLISHERS
Introduction
In coordinate geometry, understanding the concept of distance is crucial for analyzing the spatial relationships between geometric elements. The two important cases of distance measurement include:
- Distance between a point and a line
- Distance between two parallel lines
Both these concepts are frequently used in coordinate geometry problems and have significant applications in CBSE Board and JEE Mains & Advanced exams.
Table of Contents
Distance Between Point and Line Derivation
The general equation of a line is given by Ax + By + C = 0.
Consider a line L : Ax + By + C = 0 whose distance from the point P (x1, y1) is d.
Draw a perpendicular PM from the point P to the line L, as shown in the figure below.
Let Q and R be the points where the line meets the x-and y-axes, respectively.
Coordinates of the points can be written as Q(-C/A, 0) and R(0, -C/B).
Let’s derive the formula using the formulas of the area of the triangle in different aspects.
In triangle PQR,
ar(ΔPQR) = (1/2) × Base × Height = (1/2) × PM × QR
⇒ PM = [2 ar(ΔPQR)]/2….(i)
In coordinate geometry, the area of triangle with vertices (x1, y1), (x2, y2) and (x3, y3) is
Δ = (1/2) |x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)|
Here,
(x1, y1) = P(x1, y1)
(x2, y2) = Q(-C/A, 0)
(x3, y3) = R(0, -C/B)
Now, by substituting these values in area formula, we get;
ar(ΔPQR) = (1/2) |x1(0 + C/B) + (-C/A)(-C/B – y1) + 0(y1 – 0)|
ar(ΔPQR) = (1/2) |x1(C/B) + (C2/AB) + y1(C/A)|
⇒ 2 ar(ΔPQR) = |C/AB| . |Ax1 + By1 + C|….(ii)
Distance QR = √[(0 + C/A)2 + (C/B – 0)2]
Distance QR = √[(C2/A2) + (C2/B2)]
Let’s find QR using the distance formula.
QR = |C/AB| √(A2 + B2)….(iii)
Substituting (ii) and (iii) in (i),
PM = [|C/AB| . |Ax1 + By1 + C|] / [|C/AB| √(A2 + B2)]
d = |Ax1 + By1 + C|]/ √(A2 + B2)
Therefore, the perpendicular distance (d) of a line Ax + By+ C = 0 from a point (x1, y1) is given by
d = |Ax1 + By1 + C|]/ √(A2 + B2)
Based on this formula, we can derive the formula for the distance between two parallel lines.
Distance Between Two Parallel Lines
The distance between two parallel lines is equal to the perpendicular distance between the two lines. We know that the slopes of two parallel lines are the same; therefore, the equation of two parallel lines can be given as:
y = mx + c1….(1)
and y = mx + c2….(2)
The point A is the intersection point of the second line on the x-axis.
The perpendicular distance would be the required distance between two parallel lines.
The distance between the point \(\begin{array}{l}A\end{array} \) and the line \(\begin{array}{l}y\end{array} \) = \(\begin{array}{l}mx ~+ ~c_2\end{array} \) can be given by using the formula:
\(\begin{array}{l}d\end{array} \) = \(\begin{array}{l}\frac{\left | Ax_{1} + By_{1} + C \right |}{\sqrt{A^{2} + B^{2}}}\end{array} \)
\(\begin{array}{l}\Rightarrow d \end{array} \) \(\begin{array}{l}= \frac{\left | (-m)(\frac{-c_{1}}{m}) – c_{2} \right |}{\sqrt{1 + m^{2}}}\end{array} \)
\(\begin{array}{l}\Rightarrow d \end{array} \) \(\begin{array}{l}= \frac{\left | c_{1} – c_{2} \right |}{\sqrt{1 + m^{2}}}\end{array} \)
Thus, we can conclude that the distance between two parallel lines is given by:
\(\begin{array}{l}d\end{array} \) = \(\begin{array}{l}\frac{|c_1 ~- ~c_2|}{√1 + m^2}\end{array} \)
If we consider the general form of the equation of straight line, and the lines are given by:
\(\begin{array}{l} L_1 ~: ~Ax~ + ~By ~+ ~C_1\end{array} \) = \(\begin{array}{l}0\end{array} \)
\(\begin{array}{l} L_2 ~: ~Ax ~+ ~By ~+ ~C_2\end{array} \) = \(\begin{array}{l}0 \end{array} \)
Then, the distance between them is given by:
\(\begin{array}{l}d\end{array} \) = \(\begin{array}{l}\frac{|C_1 ~- ~C_2|}{√A^2~ +~ B^2}\end{array} \)
Shortest Distance Between Two parallel Lines
The shortest distance between the two parallel lines can be determined using the length of the perpendicular segment between the lines. It does not matter which perpendicular line you are choosing, as long as two points are on the line.
Thus, we can now easily calculate the distance between two parallel lines and the distance between a point and a line.
Solved Examples
Example 1:
Find the distance between two lines 3x + 4y = 9 and 6x + 8y = 15.
Solution:
Given equations of lines are:
3x + 4y = 9….(i)
6x + 8y = 15 Or 3x + 4y = 15/2 ….(ii)
Let us check whether the given lines are parallel or not.
From (i),
4y = -3x + 9
y = (-¾)x + (9/4)
Here, slope = m1 = -¾
From (ii),
8y = -6x + 15
y = (-6/8)x + (15/8)
y = (-¾)x + (15/8)
Here, slope = m2 = -¾
Thus, the slope of the given lines is equal so they are parallel to each other.
Now, by comparing with the standard form of parallel lines equations, we get:
A = 3, B = 4, C1 = -9, C2 = -15/2
d = |C1 – C2|/√(A2 + B2)
= |-9 + (15/2)|/√(9 + 16)
= |-18 + 15|/2√25
= |-3|/(2 × 5)
= 3/10
Therefore, the distance between the given lines is 3/10 units.
Example 2:
The distance between two parallel lines 5x – 12y + 2 = 0 and 5x – 12y + k = 0 is given by 5/13 units. Find the value of k.
Solution:
Given equations of lines are:
5x – 12y + 2 = 0….(i)
5x – 12y + k = 0 ….(ii)
Let us check whether the given lines are parallel or not.
From (i),
12y = 5x + 2
y = (5/12)x + (2/12)
Here, slope = m1 = 5/12
From (ii),
From (i),
12y = 5x + k
y = (5/12)x + (k/12)
Here, slope = m2 = 5/12
Thus, the slope of the given lines is equal so they are parallel to each other.
Now, by comparing with the standard form of parallel lines equations, we get:
A = 5, B = -12, C1 = 2, C2 = k
d = |C1 – C2|/√(A2 + B2)
5/13 = |2 – k|/√(25 + 144)
5/13 = |2-k|/√169
5/13 = |2 – k|/13
5 = 2 – k
k = 2 – 5 = -3
Summary
1. Distance Between a Point and a Line
The shortest distance from a given point to a line is always the perpendicular distance from the point to the line. This distance is important in various applications, including shortest path calculations, reflections, and optimizations in geometry.
Concept
If a line is given in the general form, the perpendicular distance from a given point to the line can be determined using a standard formula. This allows us to find the exact separation between any point and a line.
Applications
- Finding the shortest path from a location to a road.
- Determining whether a point lies close to or far from a given line.
- Used in physics for motion and reflection problems.
2. Distance Between Two Parallel Lines
Parallel lines are two lines in the same plane that never intersect. The distance between them remains constant at all points. This measurement is crucial in geometry for spacing, alignment, and construction.
Concept
For two given parallel lines, the perpendicular distance between them is used to determine how far apart they are. This distance remains the same at every point along the lines.
Applications
- Used in urban planning for measuring the separation between roads and railway tracks.
- Essential in mechanical engineering for designing parallel components.
- Helps in geometry problems related to congruency and transformations.
Conclusion
Understanding these distance concepts is essential for solving problems in coordinate geometry. Mastering them helps students in competitive exams like JEE Mains & Advanced, where precise calculations are needed.
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