if u extend two parallel lines, does the distance between them increase? Draw and explain.
Plzz answer.
Answers
Step-by-step explanation:
yes if we will extend two Parallel Lines the distance between them will increase.
yes if we will extend two Parallel Lines the distance between them will increase.It is so because more the line more the distance.
I can't draw the line here but you can do so that draw a normal line and one more line below it now extent both the lines .what would you find? you will find that the line has increased its size and to prove it you can even measure it by your normal measuring scale.
Answer:
Distance between 2 parallel lines is the perpendicular distance from any point to one of the lines. In this article you will learn parallel lines definition , how to find distance between them and solved examples.
The graph of the function, a pair of railroad tracks or the opposite sides of a parallelogram or keys of a piano can be a few examples of parallel lines. A common property which can be found in the above examples is that the two railroad tracks never meet, the opposite sides of a parallelogram will never intersect or the piano keys are parallel to each other. This article explains the method of finding the distance between two parallel lines with appropriate examples.
Parallel Lines Definition
Parallel lines are those lines which never meet each other. When the distance between a pair of lines is the same throughout, it can be called parallel lines. It is denoted by “||”. The main criteria for any two lines to be parallel are that they have to be drawn on the same plane. They are always equidistant from each other.
The lines can be extended till infinity. The slopes of two parallel lines are equal.
How to Find the Distance Between Two Parallel Lines
The method for calculating the distance between two parallel lines is as follows:
Ensure whether the equations of the given parallel lines are in slope-intercept form (y=mx+c).
The point of interception (c1 and c2) and slope value which is common for both the lines has to be determined.
After obtaining the above values, substitute them in the slope-intercept equation to find y.
Finally, put all the above values in the distance formula to find the distance between two parallel lines.
The two parallel lines can be taken in the form
y = mx + c1 … (i)
and y = mx + c2 … (ii)
The line (i) will intersect the x-axis at point A (–c1/m, 0) as shown in the figure.

The length of the perpendicular from point A to the line (ii) is of the same length as the distance between two lines.
Therefore, the distance between the lines (i) and (ii) is
|(–m)(–c1/m) + (–c2)|/√(1 + m2) or d = |c1–c2|/√(1+m2).
Distance d between two parallel lines y = mx + c1 and y = mx + c2 is given by
d = |C1–C2|/√A2 + B2
Distance formula d = |a1x1+b1y1+c1| / √a12+b12 , where d is the distance between two parallel lines. x1 and y1 are the two intersection points of the lines with the axis in a cartesian plane, while a1 and b1 are the coefficients of variable x and y of the line.
The equation of line through which equation of the distance formula is written as
a1x+b1y+c1 = 0
Considering the following equations of 2 parallel lines, we can calculate the distance between those lines using the distance formula
ax+by+c = 0
ax+by+c1 = 0
Using above 2 equations we can conclude that
Distance formula d = |c-c1| / √a2+b2
Also read
Height and distance problems
Trigonometric equations
Solved Examples
Example 1: Find the distance of the point (4, –6) from the line 2x – 7y – 24 = 0.
Solution:
Given line is 2x – 7y – 24 = 0. …… (1)
Comparing (1) with general equation of line Ax + By + C = 0, we get
A = 2, B = –7 and C = –24.
Given point is (x1, y1) = (4, –6).
The distance of the given point from given line is d = |Ax1 + By1 + C|/√A2+B2 = 26/7.2 = 3.571
Example 2: What is the distance of the lines 2x − 3y = 4 from the point (1, 1) measured parallel to the line x+y=1?
Solution:
The slope of line x + y = 1 is -1. It makes an angle of 135∘ with the x-axis.
The equation of the line passing through (1,1) and making an angle of 135∘ is,
x − 1/ cos135o = y − 1/sin135o = r
x − 1/(- 1/√2) = y − 1/1/√2 = r
Co-ordinates of any point on this line are
(1 − r/√2, 1 + r/√2)
If this point lies on 2x − 3y = 4, then
2(1 − r/√2) − 3(1 + r/√2) = 4
r = √2.